The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 mc010-1.jpg Fe2(SO4)3 + 2Na3PO4 W hat is the theoretical yield of Fe2(SO4)3 if 20.00 g of FePO4 reacts with an excess of Na2SO4? 26.52 g 53.04 g 150.8 g 399.9 g
2 answers:
Answer: 26.52g Solution step-by-step: 1) Chemical equation: <span>2FePO4 + 3Na2SO4 → Fe2(SO4)3 + 2Na3PO4 2) Theoretical molar ratios: 2 mol FePO4 : 3 mol Na2SO4 : 1 mol Fe2(SO4)3 : 2 mol Na3PO4 3) Convert 20.00 g of FePO4 into number of moles number of moles = mass in grams / molar mass molar mass of FePO4 = 150.82 g/mol numer of moles = 20.00 g / 150.82 g/mol = 0.1326 mol FePO4 4) Proportionality 1 mol Fe2(SO4)3 x ----------------------- = ------------------------ 2 mol FePO4 0.1326 mol FePO4 Solve for x: x = (0.1326 / 2) * 1 mol Fe2(SO4)3 = 0.0663 mol Fe2(SO4)3 5) Convert 0.0663 mol Fe2(SO4)3 into grams mass in grams = number of moles * molar mass molar mass Fe2(SO4)3 = 399.88 g/mol mass = 0.0663 mol * 399.88 g/mol = 26.51 g Depending on the number of decimal digits used by one or other you might obtain 26.52 instead 26.51, that difference does not count. </span>
Answer:
26.52 g
Explanation:
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