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

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

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

answer; /// i do believe that the correct answer is (a) boiling

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