# Question #13733

Jun 15, 2017

$C {S}_{2}$ + 3 ${O}_{2} \to$ $C {O}_{2}$ + 2 $S {O}_{2}$

${m}_{C O 2} = 4.4$ $g$

${m}_{S O 2} = 12.8$ $g$

#### Explanation:

First we need to write a balanced chemical equation:

a $C {S}_{2}$ + b ${O}_{2} \to$ c $C {O}_{2}$ + d $S {O}_{2}$

If we set a = 1, c is also 1, and d is 2, since each mole of $C {S}_{2}$ will produce one mole of $C {O}_{2}$ and two moles of $S {O}_{2}$

$C {S}_{2}$ + b ${O}_{2} \to$ $C {O}_{2}$ + 2 $S {O}_{2}$

Now there are 6 moles of $O$ on the right, so we need 3 ${O}_{2}$ on the left to balance the equation:

$C {S}_{2}$ + 3 ${O}_{2} \to$ $C {O}_{2}$ + 2 $S {O}_{2}$

Now, the molar mass of $C {S}_{2}$ is 76 g/mol. We find the number of moles present:

$n = \frac{m}{M} = \frac{7.6}{76} = 0.1$ $m o l$

The balanced equation tells us that $0.1$ $m o l$ of $C {S}_{2}$ will produce $0.1$ $m o l$ of $C {O}_{2}$ and $0.2$ $m o l$ of $S {O}_{2}$

${m}_{C O 2} = n M = 0.1 \times 44 = 4.4$ $g$

${m}_{S O 2} = n M = 0.2 \times 64 = 12.8$ $g$