# What mass of sodium chloride, NaCl, is made when 200 L of chlorine gas, Cl_2, reacts with excess sodium metal?

May 21, 2017

$1044$ $g$ of $N a C l$ are produced, since there are $8.93$ mol of $C {l}_{2}$ gas and each mole of $C {l}_{2}$ produces $2$ mol of $N a C l$.

#### Explanation:

One mole of an ideal gas has a volume of $22.4$ $L$ at STP. Chlorine is not a perfectly ideal gas but can be treated as one at this temperature and pressure.

The number of moles of gas, then, is given by $n = \frac{V}{22.4} = \frac{200}{22.4} = 8.93$ $m o l$

Looking at the balanced equation, each $1$ mol of $C {l}_{2}$ yields $2$ mol of $N a C l$, so $8.93$ mol of $C {l}_{2}$ yields $2 \times 8.93 = 17.86$ mol of $N a C l$.

The molar mass of $N a C l$ is $58.44$ $g$. To find the mass of the product, we take $n = \frac{m}{M}$ and rearrange to make $m$ the product. $m = n M = 17.86 \times 58.44 = 1044$ $g$ (with some rounding)