What is the maximum number of grams of PH3 that can be formed when 6.20 g of phosphorus reacts with 4.00 g of hydrogen to form PH3?

Question

What is the maximum number of grams of PH₃ that can be formed when 6.20 g of phosphorus reacts with 4.00 g of hydrogen to form PH₃?

P₄ + 6 H₂ --> 4 PH₃

Answer 1

P₄ + 6 H₂ --> 4 PH₃

For P₄: 6.20 g P₄ x 1 mol P₄/124 g x 4 mols PH₃ / 1 mol P₄ x 34.0 g PH₃ / mol = 6.8 g PH₃ formed

For H_2: 4.00 g H₂ x 1 mol H₂ / 2.00 g x 4 mols PH₃ / 6 mols H₂ x 34.0 g PH₃ / mol = 45.3 g PH₃ formed

Since you obtain less product with the P₄, that tells you that P₄ is LIMITING and the maximum theoretical yield would be 6.80 g of PH₃

Answer 2

6.20g P₄ x (mol / 123.895g) = 0.05 mol P₄ x (4 mol PH₃ / 1 mol P₄) = 0.2 mol PH₃

4.00g H₂ x (mol / 2.016g) = 1.98 mol H₂ x (4 mol PH3 / 6 mol H₂) = 1.32 mol PH₃

P₄ --> limiting reactant

0.2 mol PH₃ x (33.997g / mol) = 6.80g PH₃

Answer:

6.8g PH₃