Stoichiometry problem

Question

What mass of P4O10 will be obtained from the reaction of 1.33 g P4 and 5.07 g O2? What mass of P4O6 will be obtained from the reaction of 4.07 g P4 and 2.01 g O2?

Answer 1

Chemical Equation: P₄ + 5O₂ = P₄O₁₀

1.33g P₄ * (1 mol P₄ / 123.88g P₄) = 0.0107 mol P₄

5.07g O2 * (1 mol O₂ / 32g O₂) = 0.158 mol O₂

P₄ is the Limiting Reactant because 0.01.7 * 5 = 0.0536, and we will have excess O₂ from the 1:5 ratio

P₄ + 5O₂ = P₄O₁₀

I 0.0107 mol. 0.158 mol. ------

C. -0.0107 mol. -0.0536 mol +0.0107 mol

E. 0 mol 0.1048 mol 0.0107 mol

0.0107 mol P₄O₁₀ * (283.88g P₄O₁₀ / 1 mol P₄O₁₀) = 3.037516 g P₄O₁₀

Chemical Equation: P₄ + 3O₂ = P₄O6

4.07g P₄ * (1 mol P₄ / 123.88g P₄) = 0.0329 mol P₄

2.01g O2 * (1 mol O₂ / 32g O₂) = 0.0628 mol O₂

O2 is the Limiting Reactant because 0.0628 / 3 = 0.0209, and we will have excess P₄ from the 1:3 ratio

P₄ + 3O₂ = P₄O₆

I 0.0329 mol. 0.0628 mol. ------

C. -0.0209 mol. -0.0628 mol +0.0209 mol

E. 0.0119 mol 0 mol 0.0209 mol

0.0209 mol P₄O₆ * (219.88g P₄O₆ / 1 mol P₄O₆) = 4.595 g P₄O₆