Question

How many grams of Fe2O3 are formed when 16.7 moles of Fe reacts completely with

oxygen?
4Fe + 302 → 2Fe203

How many moles of H3PO4 are produced when 71.0 g P4010 reacts completely to form
H3PO4?
P4010 + 6H20 + 4H3PO4​

Answer 1

The answer to your question is 1) 1336 g of Fe₂O₃ 2) 1 mol of H₃PO₄

Step-by-step explanation:

1)

Data

mass of Fe₂O₃ = ?

moles of Fe = 16.7

Balanced chemical reaction

4Fe + 3O₂ ⇒ 2Fe₂O₃

a.- Calculate the moles of Fe₂O₃

4 moles of Fe ----------------- 2 moles of Fe₂O₃

16.7 moles of Fe -------------- x

x = (16.7 x 2)/4

x = 33.4/4

x = 8.35 moles of Fe₂O₃

b.- Calculate the molar mass of Fe₂O₃

Fe₂O₃ = (2 x 56)+ (3 x 16) = 112 + 48 = 160 g

c.- Convert moles to mass

160 g of Fe₂O₃ ------------ 1 mol

x ------------ 8.35 moles of Fe₂O₃

x = (8.35 x 160)/1

x = 1336 g of Fe₂O₃

2)

moles of H₃PO₄ = ?

mass of P₄O₁₀ = 71 g

Balanced chemical reaction

P₄O₁₀ + 6H₂O ⇒ 4H₃PO₄

a.- Calculate the molar mass of P₄O₁₀

P₄O₁₀ = (31 x 4) + (16 x 10) = 124 + 160 = 284 g

b.- Convert the mass of P₄O₁₀ to moles

284 g ------------------- 1 mol

71 g -------------------- x

x = (71 x 1)/284

x = 0.25 moles of P₄O₁₀

c.- Calculate the moles of H₃PO₄

1 mol of P₄O₁₀ ---------------- 4 moles of H₃PO₄

0.25 moles ----------------- x

x = (0.25 x 4) / 1

x = 1 mol of H₃PO₄