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 =

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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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