A mixture of CuSO4 * 5H2o and MgSO4 * 7H2O is heated until all water is lost. If

Question

A mixture of CuSO4 * 5H2o and MgSO4 * 7H2O is heated until all water is lost. If 5.020g of the mixture gives 2.988g of the anhydrous salts, what is the percent by mass of CuSO4 * 5H2O in the mixture?

Explanation / Answer

CuSO4*5H2O -----> CuSO4 + 5H2O MgSO4* 7H2O -----> MgSO4 + 7H2O mass of water = 5.020g - 2.988g = 2.032g moles of H2O = 2.032g(1mole/18g) = 0.113 mole in the mixture there are 12 moles of H2O moles of H2O from CuSO4*5H2O (5/12)*0.113 = 0.047 mole moles of H2O from MgSO4*5H2O (7/12)*0.113 = 0.066 mole since there is 1:5 mole ratio between CuSO4*5H2O&H2O then mole of CuSO4 = 0.047/5 = 0.0093 mole CuSO4*5H2O the MM of CuSO4*5H2O is 249.69 g/mol the mass of CuSO4*5H2O then is 0.0093mole*249.69 g/mol = 2.322 g the percent by mass of the CuSO4*5H2O is then (2.322g/5.020g)*100 = 46.25%

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