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 theoretically produced from 2.338 g of pure magnesium metal. Suposse magnesium metal is the limiting reactant.

Molar masses (g/mol):

AlPO4 = 121.95

Mg = 24.305

Al = 26.98

Mg3(PO4)2 = 262.86

Important note: Do not use units (i.e. g) in your answer. Use 3 digits after the decimal place in your answer.

Explanation / Answer

AlPO4(aq) + Mg(s) ---> Al(s) + Mg3(PO4)2(aq)

Balanced reaction-

2AlPO4(aq) + 3Mg(s) --> 2Al(s) + Mg3(PO4)2(aq)

As per the reaction,

3*24.305g of Mg produce = 2*26.98g of Al

1g of Mg produce = 2*26.98g/3*24.305g of Al = 0.74g of Al

So, 2.338g of Mg produce = 0.74g*2.338g of Al = 1.730g of Al

Mass of Al = 1.730g

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