When solid magnesium is reacted with an aluminum phosphate solution, metallic al

Question

When solid magnesium is reacted with an aluminum phosphate solution, metallic aluminum is produced by single displacement (equation. a): (a) AlPO4(aq)+Mg(s)Al(s)+Mg3(PO4)2(aq) Balance equation (a) and answer the following: How many grams of aluminum metal should be obtained from the reaction of 2.765 g of pure magnesium with an excess of aluminum phosphate solution, if the percent yield of the reaction is 96%.

Molar masses (g/mol):

AlPO4 = 121.95

Mg = 24.305

Al = 26.98

Mg3(PO4)2 = 262.86

Important note: Use 3 digits after the decimal place in your answer.

Explanation / Answer

Balanced chemical equation is:-

2AlPO4+6Mg=2Al+2Mg3(PO4)2

From above equation we can see that 6 moles of Mg on complete reaction produce 2 mole Al.

So 24.3×6 g Mg would produce 2×26.98g Al

2.765 g Mg would produce (53.96/145.8)×2.765

So mass of Mg produced:- 1.019 g

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