Question

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

Answer 1

`1044` `g` of `NaCl` are produced, since there are `8.93` mol of `Cl_2` gas and each mole of `Cl_2` produces `2` mol of `NaCl`.

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=V/22.4 = 200/22.4 = 8.93` `mol`

Looking at the balanced equation, each `1` mol of `Cl_2` yields `2` mol of `NaCl`, so `8.93` mol of `Cl_2` yields `2xx8.93 =17.86` mol of `NaCl`.

The molar mass of `NaCl` is `58.44` `g`. To find the mass of the product, we take `n=m/M` and rearrange to make `m` the product. `m=nM = 17.86xx58.44 = 1044` `g` (with some rounding)