Question

13. How many grams of chromium are needed to react with an excess of Cuso, to produce 27.0 g Cu?2Cr(s) + 3CuSO4 (aq) → Cr2(SO4)3(aq) + 3Cu(s)

Answer 1

In this question, we have the following reaction:

2 Cr + 3 CuSO4 -> Cr2(SO4)3 + 3 Cu

The molar ratio between Cr and Cu is 2:3, as we can see by their stoichiometric coefficients.

We have:

27.0 grams of Cu

and the molar masses are:

Cu = 63.55 g/mol

Cr = 52g/mol

Now we need to find the number of moles of Cu present in 27 grams

63.55 g = 1 mol

27 g = x moles

x = 0.425 moles of Cu in 27 grams

According to the molar ratio, 2 Cr for every 3 Cu, therefore if we have 0.425 Cu:

2 Cr = 3 Cu

x Cr = 0.425 Cu

x = 0.283 moles of Cr will be needed to produce 0.425 moles of Cu

Now with the number of moles of Cr and its molar mass, we can find the final mass:

52g = 1 mol

x grams = 0.283 moles

x = 14.7 grams are needed to react and produce 27.0 grams of Cu