Question

Calculate the amount in grams of Na2CO3 needed to react with HCL to produce 120g NaCl

Answer 1

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 :)