# Using the following reaction, if 5.93 grams of sodium cyanide are reacted with 8.45 grams of sulfuric acid, how many grams of sodium sulfate will be produced? 2NaCN + H_2SO_4 -> Na_2SO_4 + 2HCN

Jul 11, 2016

The reaction will produce 8.59 g of ${\text{Na"_2"SO}}_{4}$.

#### Explanation:

This appears to be a limiting reactant problem.

1. Write the balanced equation

$\text{2NaCN" + "H"_2"SO"_4 → "Na"_2"SO"_4 + "2HCN}$

2. Calculate the moles of ${\text{Na"_2"SO}}_{4}$ formed from the $\text{NaCN}$

${\text{Moles of Na"_2"SO"_4 = 5.93 color(red)(cancel(color(black)("g NaCN"))) × (1 color(red)(cancel(color(black)("mol NaCN"))))/(49.01 color(red)(cancel(color(black)("g NaCN")))) × (1 "mol Na"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaCN")))) = "0.060 50 mol Na"_2"SO}}_{4}$

3. Calculate the moles of ${\text{Na"_2"SO}}_{4}$ formed from the ${\text{H"_2"SO}}_{4}$

${\text{Moles of Na"_2"SO"_4 = 8.45 color(red)(cancel(color(black)("g H"_2"SO"_4))) × (1 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(98.08 color(red)(cancel(color(black)("g H"_2"SO"_4)))) × (1 "mol Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.086 15 mol Na"_2"SO}}_{4}$

4. Identify the limiting reactant

The limiting reactant is $\text{NaCN}$, because it produces the fewest moles of product.

5. Calculate the mass of ${\text{Na"_2"SO}}_{4}$

${\text{Mass of Na"_2"SO"_4 = "0.060 50" color(red)(cancel(color(black)("mol Na"_2"SO"_4))) × ("142.04 g Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "8.59 g Na"_2"SO}}_{4}$