A mixture of 4.94 g of 85.0% pure phosphine and 0.110kg of CuSO4 • 5H2O is place

Question

A mixture of 4.94 g of 85.0% pure phosphine and 0.110kg of CuSO4 • 5H2O is placed in a reaction vessel. Calculate the mass of Cu3P2 with a 6.31% yield that would be produced in the reaction: 3 CuSO4 • 5H2O + 2PH3 -> Cu3P2 + 3H2SO4 + 15H2O A mixture of 4.94 g of 85.0% pure phosphine and 0.110kg of CuSO4 • 5H2O is placed in a reaction vessel. Calculate the mass of Cu3P2 with a 6.31% yield that would be produced in the reaction: 3 CuSO4 • 5H2O + 2PH3 -> Cu3P2 + 3H2SO4 + 15H2O 3 CuSO4 • 5H2O + 2PH3 -> Cu3P2 + 3H2SO4 + 15H2O

Explanation / Answer

The mass of phosphine = (4.94 x 85) / 100 = 4.199 g

Number of moles of phosphine = 4.199 / molar mass

= 4.199 / 33.99

= 0.1235 mol

Number of moles of CuSO4.5H2O = 110 / molar mass

= 110 / 249.7

= 0.44 mol

0.44 mol of CuSO4.5H2O will require (0.44 x 2 ) / 3 = 0.293 mol of phosphine

So phosphine is the limiting reagent.

Now 0.1235 mol of phosphine will produce 0.1235/2 = 0.06175 mol of Cu3P2

So theoretical yield = 0.06175 x molar mass

= 0.06175 x 252.59

= 15.6 g

So the mass of Cu3P2 produced in the reaction = (15.6 x 6.31) / 100

= 0.98 grams

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