What's the mass of hydrogen chloride?

• I) Na2CO3 + 2 HCl ---> 2 NaCl + CO2 + H2O

Focusing on the educts: we are searching for a quantity of HCl and we know every quantity of Na2CO3. So HCl is the unknown and Na2CO3 is the known in this case.

• II) Amount of substance ratio: $\frac{searched-one (s)}{known-one (k)}$ = r ---> r = $\frac{2 mol}{1 mol}$ = 2

M(k) = $\frac{m(k)}{n(k)}$ ---> we rearrange that n(k) = $\frac{m(k)}{M(k)}$ ---> Thanks to the very first equation we know that the amount of HCl molecules is supposed to be double the amount of Na2CO3 ---> n(s) ~ r* $\frac{m(kPure)}{M(k)}$

We know the mass of Na2CO3 is not the 84,0 g, the mass is just 24% of it. So, we multiply the mass by the mass-share value w2. (0,24)

n(s) = r * $\frac{m(kPure)*w(k)}{M(k)}$ ---> n(s) = 2* $\frac{84,0 g * 0,24 }{106 g/mol}$ = 0,38 mol

• III) M(s) = $\frac{m(sPure)}{n(s)}$ ---> we rearrange that to m(sPure) = n(s) * M(s) But we know that we don't use pure HCl. So, we have to divide the amount of mass of pure HCl by the mass-share value w1. (0,18) m(s) = $\frac{n(s) * M(s)}{w1}$ ---> m(s) = $\frac{0,38 mol * 36,5 g/mol}{0,18}$ = 77,1 g

When you memorize this strategy, you will be able to solve most of the stoichiometry problems.