Question

If 12.3 mol HCl are produced in this reaction, how many grams of sodium sulfate are produced?

Answer 1

ANSWER

The mass of Na2SO4 is 874g

EXPLANATION

Given that;

The number of moles of HCl is 12.3 mol

Follow the steps below to find the mass of sodium sulfate produced

Step 1; Write a balanced equation for the reaction

$\text{ 2NaCl + H}_2SO_4\text{ }\rightarrow\text{ 2HCl + Na}_2SO_4$

In the reaction above, 2 moles of NaCl react with 1 mole of H2SO4 to give 2 moles of HCl and 1 mole of Na2SO4

Let the number of moles of Na2SO4 be x

$\begin{gathered} \text{ 2 moles HCl }\rightarrow\text{ 1 mole Na}_2SO_4 \\ \text{ 12.3 moles HCl }\rightarrow\text{ x moles Na}_2SO_4 \\ \text{ Cross multiply} \\ \text{ 2 moles HCl }\times\text{ x moles Na}_2SO_4\text{ }=\text{ 1 mole Na}_2SO_4\text{ }\times\text{ 12.3 mole HCl} \\ \text{ Isolate x} \\ \text{ }\times\text{ }=\text{ }\frac{1\text{ mole Na}_2SO_4\times12.3mol\cancel{HCl}}{2moles\cancel{HCl}} \\ \text{ } \\ \text{ x = }\frac{1\text{ }\times\text{ 12.3}}{2} \\ \text{ x = 6.15 moles} \end{gathered}$

The number of moles of Na2SO4 is 6.15 moles

Step 3; Find the mass of Na2SO4 using the below formula

$\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}$

Recall, that the molar mass of Na2SO4 is 142.04 g/mol

$\begin{gathered} \text{ mass = mole }\times\text{ molar mass} \\ \text{ mass = 6.15 }\times\text{ 142.04} \\ \text{ mass = 873.546} \\ \text{ mass = 874g Na}_2SO_4 \end{gathered}$

Therefore, the mass of Na2SO4 is 874g