Let's start off with the balanced chemical equation:
\(Na_2CO_{3_{(s)}} + \textbf2HCl_{_{(aq)}}\rightarrow CO_{2_{(g)}} + H_2O_{_{(l)}} + \textbf2NaCl_{_{(aq)}}\)
Assuming the sodium carbonate is the limiting reagent, look at the coefficients of sodium carbonate and sodium chloride and use those as a ratio of sodium carbonate to sodium chloride: 1:2.
Since you have the required mass of NaCl, convert this to moles.
Assuming you know how to find the molar mass of NaCl:
\(M_{NaCl} = 58.44g/mol\)
\(n = \frac{m}{M}\)
\(n_{NaCl} = \frac{120g}{58.44g/mol}\)
\(n_{NaCl} = 2.053mol\)
Using the ratio, since 1 mole of sodium carbonate is required to produce 2 moles of sodium chloride, cross-multiply the ratios:
\(1:2 = x:2.053mol\)
\(2x = 2.053mol\)
\(x = 1.027mol\)
Therefore 1.027 moles of sodium carbonate is required to produce the required amount of sodium chloride. Convert to mass for your final answer:
\(n_{Na_2CO_3} = 1.027mol\)
\(M_{Na_2CO_3} = 105.99g/mol\)
\(m = nM\)
\(m = (1.027mol)(105.99g/mol)\)
\(m_{Na_2CO_3} = 108.85g\)
Hope this helps :)
