Practice #6 Gram to gram Ratio Stoichiometry

Question

Copper I iodide , Cul , 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 ! 2CuSO4 +4 HI ---->

2 Cul+2H2SO4 +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

Find moles of HI: (34.56g HI) * (1 mol HI / 127.91g HI) = 0.2702 moles HI.

Then use mole ratios:

(0.2702 mol HI) * (2 mol CuI / 4 mol HI) * (190.45g / 1 mol CuI) = 25.73g CuI.

(0.2702 mol HI) * (2 mol H₂SO₄ / 4 mol HI) * (98.08g / 1 mol H₂SO₄) = 13.25g H₂SO₄.

(0.2702 mol HI) * (1 mol I₂ / 4 mol HI) * (253.8g / 1 mol I₂) = 17.14g I₂.