Balance the following chemical equation:
\qquad
SOLUTION[0]\text{H}_2 +
SOLUTION[1]\text{O}_2 \rightarrow
SOLUTION[2]\text{H}_2\text{O}
There are 2 \text{ O} on the left and only
1 on the right, so multiply
\text{H}_2\text{O} by \blue{2}.
\qquad
\text{H}_2 + \text{O}_2 \rightarrow \blue{2}\text{H}_2\text{O}
That gives us 4 \text{ H} on the right and
only 2 on the left, so multiply
\text{H}_2 by \red{2}.
\qquad
\red{2}\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}
The balanced equation is:
\qquad
2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}
\qquad
SOLUTION[0]\text{CH}_4 +
SOLUTION[1]\text{Cl}_2 \rightarrow
SOLUTION[2]\text{CCl}_4 +
SOLUTION[3]\text{HCl}
\text{C} is already balanced.
There are 4 \text{ H} on the left and only
1 on the right, so multiply
\text{HCl} by \blue{4}.
\qquad
\text{CH}_4 + \text{Cl}_2 \rightarrow \text{CCl}_4 + \blue{4}\text{HCl}
That gives us 8 \text{ Cl} on the right and
only 2 on the left, so multiply
\text{Cl}_2 by \red{4}.
\qquad
\text{CH}_4 + \red{4}\text{Cl}_2 \rightarrow \text{CCl}_4 + 4\text{HCl}
The balanced equation is:
\qquad
\text{CH}_4 + 4\text{Cl}_2 \rightarrow \text{CCl}_4 + 4\text{HCl}
\qquad
SOLUTION[0]\text{Al} +
SOLUTION[1]\text{O}_2 \rightarrow
SOLUTION[2]\text{Al}_2\text{O}_3
There are 2 \text{ O} on the left and
3 on the right. The lowest common denominator
is 6, so multiply
\text{O}_2 by \blue{3} and
\text{Al}_2\text{O}_3 by \red{2}.
\qquad
\text{Al} + \blue{3}\text{O}_2 \rightarrow \red{2}\text{Al}_2\text{O}_3
That gives us 4 \text{ Al} on the right and
only 1 on the left, so multiply
\text{Al} by \pink{4}.
\qquad
\pink{4}\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3
The balanced equation is:
\qquad
4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3
\qquad
SOLUTION[0]\text{CH}_4 +
SOLUTION[1]\text{O}_2 \rightarrow
SOLUTION[2]\text{CO}_2 +
SOLUTION[3]\text{H}_2\text{O}
\text{C} is already balanced.
There are 4 \text{ H} on the left and
2 on the right, so multiply
\text{H}_2\text{O} by \blue{2}.
\qquad
\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \blue{2}\text{H}_2\text{O}
That gives us 4 \text{ O} on the right and
only 2 on the left, so multiply
\text{O}_2 by \red{2}.
(Since oxygen is by itself on the left, it should be done
at the end because you can give it a coefficient without
affecting another element.)
\qquad
\text{CH}_4 + \red{2}\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
The balanced equation is:
\qquad
\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
\qquad
SOLUTION[0]\text{NaBr} +
SOLUTION[1]\text{Cl}_2 \rightarrow
SOLUTION[2]\text{NaCl} +
SOLUTION[3]\text{Br}_2
There is 1 \text{ Br} on the left and
2 on the right, so multiply
\text{NaBr} by \blue{2}.
\qquad
\blue{2}\text{NaBr} + \text{Cl}_2 \rightarrow \text{NaCl} + \text{Br}_2
There are 2 \text{ Cl} on the left and
1 on the right, so multiply
\text{NaCl} by \red{2}.
\qquad
2\text{NaBr} + \text{Cl}_2 \rightarrow \red{2}\text{NaCl} + \text{Br}_2
Now \text{Na} is balanced again.
The balanced equation is:
\qquad
2\text{NaBr} + \text{Cl}_2 \rightarrow 2\text{NaCl} + \text{Br}_2
\qquad
SOLUTION[0]\text{Mg} +
SOLUTION[1]\text{HCl} \rightarrow
SOLUTION[2]\text{MgCl}_2 +
SOLUTION[3]\text{H}_2
There are 2 \text{ Cl} on the right and
only 1 on the left, so multiply
\text{HCl} by \blue{2}.
\qquad
\text{Mg} + \blue{2}\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2
Now all atoms are balanced; the balanced equation is:
\qquad
\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2
\qquad
SOLUTION[0]\text{NH}_4\text{NO}_3 \rightarrow
SOLUTION[1]\text{N}_2 +
SOLUTION[2]\text{O}_2 +
SOLUTION[3]\text{H}_2\text{O}
Start with the compound that has the most elements
(\text{NH}_4\text{NO}_3).
There are 2 \text{N} on the left and
2 \text{N} on the right, so \text{N}
is already balanced.
There are 4 \text{ H} on the left and
2 on the right, so multiply
\text{H}_2\text{O} by \blue{2}.
\qquad
\text{NH}_4\text{NO}_3 \rightarrow \text{N}_2 + \text{O}_2 + \blue{2}\text{H}_2\text{O}
That gives us 3 \text{ O} on the left and
4 on the right. If we try giving
\text{O}_2 a coefficient of \red{\frac{1}{2}},
it gives us 3 \text{ O} on both sides.
\qquad
\text{NH}_4\text{NO}_3 \rightarrow \text{N}_2 + \red{\frac{1}{2}}\text{O}_2 + 2\text{H}_2\text{O}
Since fractions are not usually used as coefficients, multiply everything by 2 to get rid of the fraction.
\qquad
2\text{NH}_4\text{NO}_3 \rightarrow 2\text{N}_2 + 1\text{O}_2 + 4\text{H}_2\text{O}
The balanced equation is:
\qquad
2\text{NH}_4\text{NO}_3 \rightarrow 2\text{N}_2 + \text{O}_2 + 4\text{H}_2\text{O}
\qquad
SOLUTION[0]\text{C}_2\text{H}_6\text{O} +
SOLUTION[1]\text{O}_2 \rightarrow
SOLUTION[2]\text{CO}_2 +
SOLUTION[3]\text{H}_2\text{O}
For a combustion reaction, it is usually easiest to start
with \text{C}.
There are 2 \text{ C} on the left and
1 on the right, so multiply
\text{CO}_2 by \blue{2}.
\qquad
\text{C}_2\text{H}_6\text{O} + \text{O}_2 \rightarrow \blue{2}\text{CO}_2 + \text{H}_2\text{O}
There are 6 \text{ H} on the left and
2 on the right, so multiply
\text{H}_2\text{O} by \red{3}.
\qquad
\text{C}_2\text{H}_6\text{O} + \text{O}_2 \rightarrow 2\text{CO}_2 + \red{3}\text{H}_2\text{O}
That gives us 7 \text{ O} on the right and
3 on the left, so multiply
\text{O}_2 by \pink{3}.
\qquad
\text{C}_2\text{H}_6\text{O} + \pink{3}\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}
The balanced equation is:
\qquad
\text{C}_2\text{H}_6\text{O} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}
\qquad
SOLUTION[0]\text{Mg} +
SOLUTION[1]\text{O}_2 \rightarrow
SOLUTION[2]\text{MgO}
There are 2 \text{ O} on the left and only
1 on the right, so multiply
\text{MgO} by \blue{2}.
\qquad
\text{Mg} + \text{O}_2 \rightarrow \blue{2}\text{MgO}
That gives us 2 \text{ Mg} on the right and
only 1 on the left, so multiply
\text{Mg} by \red{2}.
\qquad
\red{2}\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}
The balanced equation is:
\qquad
2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}
\qquad
SOLUTION[0]\text{Al} +
SOLUTION[1]\text{HCl} \rightarrow
SOLUTION[2]\text{AlCl}_3 +
SOLUTION[3]\text{H}_2
There is 1 \text{ H} and 1 \text{ Cl} on the left and
3 \text{ Cl} and 2 \text{ H} on the right.
The lowest common denominator for the right is 6, so multiply
\text{AlCl}_3 by \blue{2} and
\text{H}_2 by \red{3}.
\qquad
\text{Al} + \text{HCl} \rightarrow \blue{2}\text{AlCl}_3 + \red{3}\text{H}_2
That gives us 6 \text{ H} and 6 \text{ Cl}
on the right, so multiply
\text{HCl} by \pink{6}.
\qquad
\text{Al} + \pink{6}\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2
That gives us 2 \text{ Al} on the right and
only 1 on the left, so multiply
\text{Al} by \green{2}.
\qquad
\green{2}\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2
The balanced equation is:
\qquad
2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2
\qquad
SOLUTION[0]\text{CaCl}_2 +
SOLUTION[1]\text{Na}_3\text{PO}_4 \rightarrow
SOLUTION[2]\text{Ca}_3\text{(PO}_4\text{)}_2 +
SOLUTION[3]\text{NaCl}
We can treat the phosphate polyatomic ion \text{(PO}_4\text{)} as an atom, symbolized by \green{X}:
\qquad
\text{CaCl}_2 + \text{Na}_3\green{\text{PO}_4} \rightarrow \text{Ca}_3(\green{\text{PO}_4\text{}})_2 + \text{NaCl}
\qquad
\text{CaCl}_2 + \text{Na}_3\green{X} \rightarrow \text{Ca}_3\green{X}_2 + \text{NaCl}
There is 1 \space X on the left and 2 \space X
on the right, so multiply
\text{Na}_3X by \blue{2}.
\qquad
\text{CaCl}_2 + \blue{2}\text{Na}_3X \rightarrow \text{Ca}_3X_2 + \text{NaCl}
There are 6 \text{ Na} on the left and only
1 on the right, so multiply
\text{NaCl} by \red{6}.
\qquad
\text{CaCl}_2 + 2\text{Na}_3X \rightarrow \text{Ca}_3X_2 + \red{6}\text{NaCl}
That gives us 6 \text{ Cl} on the right and
only 2 on the left, so multiply
\text{CaCl}_2 by \pink{3}.
\qquad
\pink{3}\text{CaCl}_2 + 2\text{Na}_3X \rightarrow \text{Ca}_3X_2 + 6\text{NaCl}
Now \text{Ca} is balanced too.
Replacing \text{PO}_4 for X, the balanced equation is:
\qquad
3\text{CaCl}_2 + 2\text{Na}_3\text{PO}_4 \rightarrow \text{Ca}_3\text{(PO}_4\text{)}_2 + 6\text{NaCl}
\qquad
SOLUTION[0]\text{N}_2\text{H}_4 +
SOLUTION[1]\text{O}_2 \rightarrow
SOLUTION[2]\text{NO}_2 +
SOLUTION[3]\text{H}_2\text{O}
There are 2 \text{ N} on the left and
only 1 on the right, so multiply
\text{NO}_2 by \blue{2}.
\qquad
\text{N}_2\text{H}_4 + \text{O}_2 \rightarrow \blue{2}\text{NO}_2 + \text{H}_2\text{O}
There are 4 \text{ H} on the left and
only 2 on the right, so multiply
\text{H}_2\text{O} by \red{2}.
\qquad
\text{N}_2\text{H}_4 + \text{O}_2 \rightarrow 2\text{NO}_2 + \red{2}\text{H}_2\text{O}
That gives us 6 \text{ O} on the right and
only 2 on the left, so multiply
\text{O}_2 by \pink{3}.
(Since oxygen is by itself on the left, it should be done
at the end because you can give it a coefficient without
affecting another element.)
\qquad
\text{N}_2\text{H}_4 + \pink{3}\text{O}_2 \rightarrow 2\text{NO}_2 + 2\text{H}_2\text{O}
The balanced equation is:
\qquad
\text{N}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{NO}_2 + 2\text{H}_2\text{O}
\qquad
SOLUTION[0]\text{Fe} +
SOLUTION[1]\text{H}_2\text{O} \rightarrow
SOLUTION[2]\text{Fe}_3\text{O}_4 +
SOLUTION[3]\text{H}_2
There are 3 \text{ Fe} on the right and
only 1 on the left, so multiply
\text{Fe} by \blue{3}.
\qquad
\blue{3}\text{Fe} + \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2
There are 4 \text{ O} on the right and
only 1 on the left, so multiply
\text{H}_2\text{O} by \red{4}.
\qquad
3\text{Fe} + \red{4}\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2
That gives us 8 \text{ H} on the left and
only 2 on the right, so multiply
\text{H}_2 by \pink{4}.
(Since hydrogen is by itself on the right, it should be done
at the end because you can give it a coefficient without
affecting another element.)
\qquad
3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \pink{4}\text{H}_2
The balanced equation is:
\qquad
3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2
\qquad
SOLUTION[0]\text{NH}_4\text{NO}_3 \rightarrow
SOLUTION[1]\text{N}_2\text{O} +
SOLUTION[2]\text{H}_2\text{O}
There are 2 \text{ N} on the right and
2 on the left, so \text{N}
is already balanced.
There are 4 \text{ H} on the left and
only 2 on the right, so multiply
\text{H}_2\text{O} by \blue{2}.
\qquad
\text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + \blue{2}\text{H}_2\text{O}
There are 3 \text{ O} on the right and
3 on the left, so \text{O}
is already balanced.
The balanced equation is:
\qquad
\text{NH}_4\text{NO}_3 \rightarrow \text{N}_2\text{O} + 2\text{H}_2\text{O}
\qquad
SOLUTION[0]\text{HgO} \rightarrow
SOLUTION[1]\text{Hg} +
SOLUTION[2]\text{O}_2
There are 2 \text{ O} on the right and
only 1 on the left, so multiply
\text{HgO} by \blue{2}.
\qquad
\blue{2}\text{HgO} \rightarrow \text{Hg} + \text{O}_2
Now there are 2 \text{ Hg} on the left and
only 1 on the right, so multiply
\text{Hg} by \red{2}.
\qquad
2\text{HgO} \rightarrow \red{2}\text{Hg} + \text{O}_2
The balanced equation is:
\qquad
2\text{HgO} \rightarrow 2\text{Hg} + \text{O}_2
\qquad
SOLUTION[0]\text{SiO}_2 +
SOLUTION[1]\text{HF} \rightarrow
SOLUTION[2]\text{SiF}_4 +
SOLUTION[3]\text{H}_2\text{O}
There is 1 \text{ Si} on the right and
1 on the left, so \text{Si}
is already balanced.
There are 2 \text{ O} on the left and
only 1 on the right, so multiply
\text{H}_2\text{O} by \blue{2}.
\qquad
\text{SiO}_2 + \text{HF} \rightarrow \text{SiF}_4 + \blue{2}\text{H}_2\text{O}
Now there are 4 \text{ H} on the right and
only 1 on the left, so multiply
\text{HF} by \red{4}.
\qquad
\text{SiO}_2 + \red{4}\text{HF} \rightarrow \text{SiF}_4 + 2\text{H}_2\text{O}
Now \text{F} is balanced too.
The balanced equation is:
\qquad
\text{SiO}_2 + 4\text{HF} \rightarrow \text{SiF}_4 + 2\text{H}_2\text{O}
\qquad
SOLUTION[0]\text{Mg(OH)}_2 +
SOLUTION[1]\text{HCl} \rightarrow
SOLUTION[2]\text{MgCl}_2 +
SOLUTION[3]\text{H}_2\text{O}
There is 1 \text{ Mg} on the right and
1 on the left, so \text{Mg}
is already balanced.
There are 2 \text{ Cl} on the right and
only 1 on the left, so multiply
\text{HCl} by \blue{2}.
\qquad
\text{Mg(OH)}_2 + \blue{2}\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\text{O}
Now there are 4 \text{ H} on the left and
only 2 on the right, so multiply
\text{H}_2\text{O} by \red{2}.
\qquad
\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \red{2}\text{H}_2\text{O}
Now \text{O} is balanced too.
The balanced equation is:
\qquad
\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}
\qquad
SOLUTION[0]\text{H}_2\text{SO}_4 +
SOLUTION[1]\text{Pb(OH)}_4 \rightarrow
SOLUTION[2]\text{Pb(SO}_4\text{)}_2 +
SOLUTION[3]\text{H}_2\text{O}
We can treat the sulfate polyatomic ion \text{(SO}_4\text{)} as an atom, symbolized by \green{X}:
\qquad
\text{H}_2\green{\text{SO}_4} + \text{Pb(OH)}_4 \rightarrow \text{Pb(}\green{\text{SO}_4}\text{)}_2 + \text{H}_2\text{O}
\qquad
\text{H}_2\green{X} + \text{Pb(OH)}_4 \rightarrow \text{Pb}\green{X}_2 + \text{H}_2\text{O}
There is 1 \space X on the left and 2 \space X
on the right, so multiply
\text{H}_2X by \blue{2}.
\qquad
\blue{2}\text{H}_2X + \text{Pb(OH)}_4 \rightarrow \text{Pb}X_2 + \text{H}_2\text{O}
That gives us 8 \text{ H} on the left and
only 2 on the right, so multiply
\text{H}_2\text{O} by \red{4}.
\qquad
2\text{H}_2X + \text{Pb(OH)}_4 \rightarrow \text{Pb}X_2 + \red{4}\text{H}_2\text{O}
Everything is now balanced. Replacing \text{SO}_4 for X, the balanced equation is:
\qquad
2\text{H}_2\text{SO}_4 + \text{Pb(OH)}_4 \rightarrow \text{Pb(}\text{SO}_4\text{)}_2 + 4\text{H}_2\text{O}
\qquad
SOLUTION[0]\text{As}_4\text{S}_6 +
SOLUTION[1]\text{O}_2 \rightarrow
SOLUTION[2]\text{As}_4\text{O}_6 +
SOLUTION[3]\text{SO}_2
\text{As} is already balanced.
There are 6 \text{ S} on the left and
only 1 on the right, so multiply
\text{SO}_2 by \blue{6}.
\qquad
\text{As}_4\text{S}_6 + \text{O}_2 \rightarrow \text{As}_4\text{O}_6 + \blue{6}\text{SO}_2
That gives us 18 \text{ O} on the right and
only 2 on the left, so multiply
\text{O}_2 by \red{9}.
(Since oxygen is by itself on the left, it should be done
at the end because you can give it a coefficient without
affecting another element.)
\qquad
\text{As}_4\text{S}_6 + \red{9}\text{O}_2 \rightarrow \text{As}_4\text{O}_6 + 6\text{SO}_2
The balanced equation is:
\qquad
\text{As}_4\text{S}_6 + 9\text{O}_2 \rightarrow \text{As}_4\text{O}_6 + 6\text{SO}_2
\qquad
SOLUTION[0]\text{Cr}_2\text{O}_3 +
SOLUTION[1]\text{Mg} \rightarrow
SOLUTION[2]\text{Cr} +
SOLUTION[3]\text{MgO}
There are 2 \text{ Cr} on the left and
only 1 on the right, so multiply
\text{Cr} by \blue{2}.
\qquad
\text{Cr}_2\text{O}_3 + \text{Mg} \rightarrow \blue{2}\text{Cr} + \text{MgO}
There are 3 \text{ O} on the left and
only 1 on the right, so multiply
\text{MgO} by \red{3}.
\qquad
\text{Cr}_2\text{O}_3 + \text{Mg} \rightarrow 2\text{Cr} + \red{3}\text{MgO}
That gives us 3 \text{ Mg} on the right and
only 1 on the left, so multiply
\text{Mg} by \pink{3}.
\qquad
\text{Cr}_2\text{O}_3 + \pink{3}\text{Mg} \rightarrow 2\text{Cr} + 3\text{MgO}
The balanced equation is:
\qquad
\text{Cr}_2\text{O}_3 + 3\text{Mg} \rightarrow 2\text{Cr} + 3\text{MgO}