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If 32.73 g of FePO4, react with excess Na2SO4, how many grams of Fe2(SO4)3, can be made? (Sig Figs Count)

If 32.73 g of FePO4, react with excess Na2SO4, how many grams of Fe2(SO4)3, can be-example-1
User Dliix
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1 Answer

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Step-by-step explanation:

First, we need to transform 32.73 g of FePO4 into moles.

For this, we use the following formula: moles = mass/molar mass

molar mass of FePO4 = 150.82 g/mol

moles = 32.73/150.82

moles = 0.22 moles

Now we use the equation ratio between FePO4 and Fe2(SO4)3 to find out the quantity in moles of Fe2(SO4)3.

2FePO4 + 3Na2SO4 → Fe2(SO4)3 + 2Na3PO4

2 moles of FePO4 produces 1 moles of Fe2(SO4)3

So:

2 moles of FePO4 --- 1 moles of Fe2(SO4)3

0.22 moles of FePO4 --- x moles of Fe2(SO4)3

x = 0.11 moles of Fe2(SO4)3

Now we transform the quantity of moles of Fe2(SO4)3 into grams using the following formula: mass = moles x molar mass

molar mass of Fe2(SO4)3 = 399.88 g/mol

mass = 0.11 x 399.88

mass = 43.39 g

Answer: mass = 43.39 g

User Hamed MP
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