Asked by jeff
In the explosion of a hydrogen-filled balloon, 0.54 g of hydrogen reacted with 6.6 g of oxygen to form how many grams of water vapor?
Answers
Answered by
bobpursley
Figure the moles of each.
.54/2=.27 moles H2
6.6/32=.21moles O2
Balance the reaction:
2H2+O2>>2H2O
well, you need twice as much H2 as O2, so H2 is the limiting reactant. Use the limiting reactant.
.27molesH2 (2moleswater/2molesH2)=.27moleswater
figure the mass of .27 moles water
.54/2=.27 moles H2
6.6/32=.21moles O2
Balance the reaction:
2H2+O2>>2H2O
well, you need twice as much H2 as O2, so H2 is the limiting reactant. Use the limiting reactant.
.27molesH2 (2moleswater/2molesH2)=.27moleswater
figure the mass of .27 moles water
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