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Consider the reaction3Mg (s)+2H3PO4(aq)=Mg3(PO4)2(s)+3H2(g)How many grams of magnesium phosphate (MW = 262.86 g/mol) should be produced if 10.800 g of magnesium react with excess phosphoric acid? Give your answer with three significant digits.Give your answer with three 3 significant figures



Answer :

Answer

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Explanation

Given:

[tex]3Mg_{(s)}+\text{ 2H}_3PO_{4(aq)}\rightarrow Mg_3(PO_4)_{2(s)}+3H_{2(g)}[/tex]

Mass of Mg = 10.800 g

Molar mass of Magnesium phosphate = 262.86 g/mol

Required: Mass of Magnesium phosphate.

Solution:

Step 1: Calculate the number of moles of magnesium

n = m/M where n is the moles, m is the mass and M is the molar mass

n = 10.800g/24.305g/mol

n = 0.444 mol

Step 2: Use the stoichiometry to calculate the moles of Mg3(PO4)2

The molar ratio between Mg and M3(PO4)2 is 3:1

Therefore the number of moles of Mg3(PO4)2 = 0.444 mol x (1/3) = 0.148 mol

Step 3: Use the moles to calculate the mass of Mg3(PO4)2

m = n x M

m = 0.148 mol x 262.86 g/mol

m =

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