Limiting reactants in stoichiometry and percent yield.
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Use the following equation:
H2SO4 + Al produces Al2(SO4)3 + H2
Given this information: 350.00 mL of 6.00 M H2SO4 reacts with 10.00 grams of Al. This collects 13.55 L of H2. What is the percent yield of the reaction?
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Solution Summary
The ideal gas equation states that pV=nRT.
p=pressure
V=volume
n=number of moles
R=ideal gas constant
T=Temperature
Solution Preview
In solving this problem, the first thing you need to do is balance the presented equation, such that there are equimolar quantities of reagents and products, thus:
H2SO4 + Al produces Al2(SO4)3 + H2
is balanced to give:
3H2SO4 + 2Al produces Al2(SO4)3 + 3H2
since the product Al2(SO4)3 contains 2xAl and 3xSO4 (ie 3xH2SO4 are required). In turn, 3xH2SO4 will produce 3xH2.
If we use 350ml of 6M H2SO4, the number of moles of H2SO4 is equal ...
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