Question

Question 18 (of 18) 5.65 points Ch. Ex. 55- Calculate Mass Formed Using Ideal Gas Law 1 out of 3 attempts Enter your answer in the provided box. Assistance How many grams of phosphine (PHy) can form when 253 g of phosphorus and 86.1 L of hydrogen gas react at STP? P-Be PBye) abalanced] g PH

Answer 1

If you balanced this equation, then equation will be P4(s) + 2H2(g) =2 PH3. So one mole of P4 produced 2 moles of PH3. Molecular weight of P4 is 123.895 g. Here P4 used is 25.5 g. So (25.5÷123.895) mole = 0.2058 mole P4 is used. Since 1 mole P4 produced 2 moles of PH3, so 0.2058 mole P4 will produced (0.2058 × 2) mole = 0.4116 mole of PH3. Molecular weight of PH3 is 33.9975 g/mol. So 1 mole PH3 = 33.9975 g PH3. So 0.4116 mole of PH3 = (0.4116×33.9975) g of PH3 = 13.993 g of PH3.

So answer is 13.993 g of PH3. Nearly 14 g of PH3.