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5. Given the following equation: 2 FePO4+3 Na2SO4 → Fe2(SO4)3 + 2 Na3PO4 If I perform this reaction with 25 g of Iron (III) phosphate (FePO4) and 33 g of sodium sulfate(Na₂SO4), what is the limiting and excess reactant and how much sodium phosphate (Na,PO4) will be formed? Limiting Reactant: Excess Reactant: Part 1: 25g FePO4 Step 1: mass to moles Molar mass of FePO4 =150.82 g/mol 25g FePO4 + molar mass of FePO4 = moles FePO4 Step 2: mole conversion from FePO4 to Na3PO4 using the coefficients from the balanced equation mol mol X = mol Na3PO4 mol Step 3: Convert from moles of Na3PO4 to grams of Na3PO4 by multiplying the moles in step 2 by its molar mass. Molar mass of Na3PO4 = g/mol Step 2 moles Na3PO4 x molar mass of Na3PO4 g = g Na3PO4 Part 2: 33g Na₂SO4 Step 1: mass to moles Molar mass of Na₂SO4 = g/mol 33g Na₂SO4 + molar mass of Na₂SO4 = moles Na₂SO4 Step 2: mole conversion from Na₂SO4 to Na,PO4 using the coefficients from the balanced equation mol mol mol Na3PO4 X mol Step 3: Convert from moles of Na-PO4 to grams of Na-PO, by multiplying the moles in step 2 by its molar mass. g Na3PO4 Step 2 moles Na3PO4 x molar mass of Na3PO4 g =

5. Given the following equation: 2 FePO4+3 Na2SO4 → Fe2(SO4)3 + 2 Na3PO4 If I perform this reaction with 25 g of Iron (III) phosphate (FePO4) and 33 g of sodium sulfate(Na₂SO4), what is the limiting and excess reactant and how much sodium phosphate (Na,PO4) will be formed? Limiting Reactant: Excess Reactant: Part 1: 25g FePO4 Step 1: mass to moles Molar mass of FePO4 =150.82 g/mol 25g FePO4 + molar mass of FePO4 = moles FePO4 Step 2: mole conversion from FePO4 to Na3PO4 using the coefficients from the balanced equation mol mol X = mol Na3PO4 mol Step 3: Convert from moles of Na3PO4 to grams of Na3PO4 by multiplying the moles in step 2 by its molar mass. Molar mass of Na3PO4 = g/mol Step 2 moles Na3PO4 x molar mass of Na3PO4 g = g Na3PO4 Part 2: 33g Na₂SO4 Step 1: mass to moles Molar mass of Na₂SO4 = g/mol 33g Na₂SO4 + molar mass of Na₂SO4 = moles Na₂SO4 Step 2: mole conversion from Na₂SO4 to Na,PO4 using the coefficients from the balanced equation mol mol mol Na3PO4 X mol Step 3: Convert from moles of Na-PO4 to grams of Na-PO, by multiplying the moles in step 2 by its molar mass. g Na3PO4 Step 2 moles Na3PO4 x molar mass of Na3PO4 g =

Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter9: Chemical Quantities
Section: Chapter Questions
Problem 15ALQ: Consider the balanced chemical equation :math>A+5B3C+4D en equal masses of A and B are reacted,...
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5. Given the following equation: 2 FePO4+3 Na2SO4 → Fe2(SO4)3 + 2 Na3PO4 If I perform this reaction with 25 g of Iron (III) phosphate (FePO4) and 33 g of sodium sulfate(Na₂SO4), what is the limiting and excess reactant and how much sodium phosphate (Na,PO) will be formed? Limiting Reactant: Excess Reactant: Part 1: 25g FePO4 Step 1: mass to moles Molar mass of FePO4 =150.82 g/mol 25g FePO4 + molar mass of FePO4 = moles FePO4 Step 2: mole conversion from FePO4 to Na PO, using the coefficients from the balanced equation mol mol X = mol Na3PO4 mol Step 3: Convert from moles of Na3PO4 to grams of Na3PO4 by multiplying the moles in step 2 by its molar mass. Molar mass of Na3PO4 = g/mol Step 2 moles Na3PO4 x molar mass of Na3PO4 g = g Na3PO4 Part 2: 33g Na₂SO4 Step 1: mass to moles Molar mass of Na₂SO4 = g/mol moles Na₂SO4 33g Na₂SO4 + molar mass of Na₂SO4 = Step 2: mole conversion from Na₂SO4 to Na,PO4 using the coefficients from the balanced equation mol mol Na3PO4 mol X mol Step 3: Convert from moles of Na-PO4 to grams of Na-PO4 by multiplying the moles in step 2 by its molar mass. g Na3PO4 Step 2 moles Na3PO4 x molar mass of Na3PO4 g =
Transcribed Image Text:5. Given the following equation: 2 FePO4+3 Na2SO4 → Fe2(SO4)3 + 2 Na3PO4 If I perform this reaction with 25 g of Iron (III) phosphate (FePO4) and 33 g of sodium sulfate(Na₂SO4), what is the limiting and excess reactant and how much sodium phosphate (Na,PO) will be formed? Limiting Reactant: Excess Reactant: Part 1: 25g FePO4 Step 1: mass to moles Molar mass of FePO4 =150.82 g/mol 25g FePO4 + molar mass of FePO4 = moles FePO4 Step 2: mole conversion from FePO4 to Na PO, using the coefficients from the balanced equation mol mol X = mol Na3PO4 mol Step 3: Convert from moles of Na3PO4 to grams of Na3PO4 by multiplying the moles in step 2 by its molar mass. Molar mass of Na3PO4 = g/mol Step 2 moles Na3PO4 x molar mass of Na3PO4 g = g Na3PO4 Part 2: 33g Na₂SO4 Step 1: mass to moles Molar mass of Na₂SO4 = g/mol moles Na₂SO4 33g Na₂SO4 + molar mass of Na₂SO4 = Step 2: mole conversion from Na₂SO4 to Na,PO4 using the coefficients from the balanced equation mol mol Na3PO4 mol X mol Step 3: Convert from moles of Na-PO4 to grams of Na-PO4 by multiplying the moles in step 2 by its molar mass. g Na3PO4 Step 2 moles Na3PO4 x molar mass of Na3PO4 g =
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