Question

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 1

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Step-by-step explanation

Given:

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

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 =