Question

How many moles of NaCl will react completely with 18.5 L F2 gas at 300.0 K and 1.00 atm? 0.751 mol NaCl 1.33 mol NaCl 1.50 mol NaCl 2.66 mol NaCl

Answer 1

`1.50` moles of `NaCl`.

Explanation:

This is our balanced single-substitution reaction:

`2NaCl + F_2 -> 2NaF + Cl_2`

From this, we can see that `1` mole of `F_2` gas corresponds to `2` moles of `NaCl`.

To calculate the moles of `NaCl` needed to react with `F_2` gas, we'll first need to find the number of moles of `F_2`.
This can be done using the Ideal Gas Law:

`pV = nRT`

Where `p` is pressure, `V` is volume, `n` is number of moles, `R` is the gas constant, and `T` is temperature.

We're already given volume, temperature, and pressure from the question, so we can just plug those values in and solve for `n`:

`pV = nRT`

`n = (pV)/(RT) = ("1.00 atm" xx "18.5 L")/("0.08206 L atm/K mol" xx "300.0 K") = "0.751 mol"`

Since `1` mole of `F_2` gas corresponds to `2` moles of `NaCl`, `"0.751"` moles of `F_2` gas will correspond to `2xx0.751 = 1.50` moles of `NaCl`.