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 underlying chemical equation here is

2 CuSO₄ + 4 HI → 2 CuI + 2 H₂​SO₄ + I₂

which you can use to determine the mole ratios between different reactants and products in the calculations

we're given that 34.56 grams​ of HI interacts, so you want to start by converting that over into moles by dividing by the Molar Mass

Moles of HI = 34.56 g / (127.912 g/mol) = 0.2702 mol HI

from there then you can use the molar ratios and molar masses of each product to find the grams produced

0.2702 mol HI x 2 mol CuI/4 mol HI x 190.450 g / mol = 25.73 g CuI

0.2702 mol HI x 2 mol H₂SO₄ / 4 mol HI x 98.076 g H₂SO₄ = 13.25 g H₂SO₄

0.2702 mol HI x 1 mol I₂ / 4 mol HI x 253.808 g I₂ = 17.14 g I₂

Answer 2

25.73 g CuI

17.14 g I₂