Practice Question #6 – Gram to Gram Mole Ratio Stoichiometry

Question

1.     Copper I iodide, CuI, is not stable enough to last long in storage, so it is generally made just prior to use. In can be prepared from copper sulphate and hydriodic acid by the following reaction. SHOW WORK!

2 CuSO₄ + 4 HI -------> 2 CuI + 2 H₂SO₄ + I₂

If 34.56 grams of HI is used, calculate the number of grams of each of the products that are produced.

Answer 1

The first thing to do is to balance the chemical equation. This is already completed for us. This provides the mole ratios of each component in the reaction. Once we have this, we can interconvert between any specie in the reaction.

Calculation of grams of CuI:

34.56 g HI * (1 mol HI / 127.904 g HI) * (2 mol CuI / 4 mol HI) * (190.45 g CuI / 1 mol CuI) = 25.73 g CuI

Calculation of grams of H2SO4:

34.56 g HI * (1 mol HI / 127.904 g HI) * (2 mol H2SO4 / 4 mol HI) * (98.08 g H2SO4 / 1 mol H2SO4) = 13.25 g H2SO4

Calculation of grams of I2:

34.56 g HI * (1 mol HI / 127.904 g HI) * (1 mol I2 / 4 mol HI) * (253.809 g I2 / 1 mol I2) = 17.14 g I2

In each case, the first step is to convert into moles of HI. This is done by dividing by the molar mass (grams/mole) of HI. Then, we use the mole ratio that is given from the coefficients of the balanced chemical equation to interconvert between reagents. This is the number in front of each specie. Finally, we conver the final product into grams by multiplying by its respective molar mass.