I need to know the answer to these problems. Please, please help me!!!! THANKS SO MUCH!!!!!!!!
1. The reaction below was known to proceed at 85.3% yield of SO2.
P4S5(s) + 5O2(g)--> P4O10(s) + 5SO2(g)
(Molar mass P4S5 = 284.2 g/mol)
How many grams of P4S5 must be burned in excess oxygen to produce 50.0 grams of SO2
2.
2006-09-22 11:15:32 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
MM SO2 = 64.06 g/mol
mass of P4S5 =
50g SO2*(1mol/64.06g)SO2*(1molP4S5/5molSO2)*(284.2g/mol)
= 44.4 g
best answer?
2006-09-22 23:18:20 · answer #1 · answered by teroy 4 · 0⤊ 0⤋
The answer is 44.4 grams of P4S5. This is a relatively easy question! You don't need to find the limiting reactant because the question already indicated that oxygen was in excess making P4S5 the limiting reactant. However you still need to balance the equation. Your balanced equation is
P4S5 +10O2 ----> P4O10 + 5SO2
from the equation 1 mole of P4S5 yields 5 moles of SO2
then convert 5 moles of SO2 to grams by multiplying the molecula mass of SO2 by 5, that gives 320 g
This means that 1 mole or 284.4 g of P4S5 yields 320g of SO2
what you need to do next is find how many grams of P4S5 will produce 50g of SO2 by multiplying
284.2 g of P4S5 by 50g of SO2, and all divided by 320g of SO2
If you did that right your answer should be 44.4 grams of P4S5. PEACE!
2006-09-22 11:57:08 · answer #2 · answered by scooby doo 3 · 0⤊ 0⤋
actually it is 52g because you first need to take into account that 85.3% of SO2 is being produced; 50g/0.853=58.62g. then you continue to find the moles and then the reactants moles (convert to grams) to find the answer
2016-01-13 07:39:36 · answer #3 · answered by ? 1 · 0⤊ 0⤋
try paying attention in class
2006-09-22 11:26:08 · answer #4 · answered by Anonymous · 0⤊ 0⤋
51.96g P4S5
2006-09-22 11:26:52 · answer #5 · answered by Greg G 5 · 1⤊ 0⤋
48.022
refer ur textbook for explanation
2006-09-22 11:20:58 · answer #6 · answered by ashwin_hariharan 3 · 0⤊ 0⤋
6grams
2006-09-22 11:36:22 · answer #7 · answered by kellem101 2 · 0⤊ 0⤋