Question

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`

Answer 1

The reaction will produce 8.59 g of `"Na"_2"SO"_4`.

Explanation:

This appears to be a limiting reactant problem.

1. Write the balanced equation

`"2NaCN" + "H"_2"SO"_4 → "Na"_2"SO"_4 + "2HCN"`

2. Calculate the moles of `"Na"_2"SO"_4` formed from the `"NaCN"`

`"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 `"Na"_2"SO"_4` formed from the `"H"_2"SO"_4`

`"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 `"NaCN"`, because it produces the fewest moles of product.

5. Calculate the mass of `"Na"_2"SO"_4`

`"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`