Asked by jecce.
If 10.0 g of hydrogen gar reacts with 20.0 g of oxygen gas, how much water will be made?
So far this is all i have:
(10.0g H2/1)(2molH2/2.02H2) = 9.0099 mol H2
(20.0g O2/1)(2mol O2/32g O2) = 1.25 mol O2
With the limited reactant being O2
Now what do i do?
I also have to find the yield of only 20.0g of H20 that was made.
So far this is all i have:
(10.0g H2/1)(2molH2/2.02H2) = 9.0099 mol H2
(20.0g O2/1)(2mol O2/32g O2) = 1.25 mol O2
With the limited reactant being O2
Now what do i do?
I also have to find the yield of only 20.0g of H20 that was made.
Answers
Answered by
bobpursley
Recheck your math onthe moles of H2.
Yes, the limiting reactant was O2, it has less moles.
H2 + 1/2 02>>H2O
so, you only have 1.25moles O2, so you can use up to 2.50 moles H2, and you will get 2.5 moles water.
Yes, the limiting reactant was O2, it has less moles.
H2 + 1/2 02>>H2O
so, you only have 1.25moles O2, so you can use up to 2.50 moles H2, and you will get 2.5 moles water.
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