Question

WHAT is the percent yield of H2O if 87.0 g of H2O is produced by combining 95.0 g of O2 and 11.0 g of H2

Answer 1

The correct answer is letter c 88.5%. The percent yield of H2O if 87.0 g of H2O is produced by combining 95.0 g of O2 and 11.0g of H2 is 88.5%

Here are the choices.
a 56.5% 
b 59.0% 
c 88.5% 
d 99.7%

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Answer 2

Answer:The percent yield of water is 87.87%.

Explanation:

Experimental yield of water =  87.0 g

Theoretical yield of water will be calculate as:

$O_2+2H_2\rightarrow 2H_2O$

Moles of oxygen gas :

$\frac{95.0g}{32 g/mol}=2.96 moles$

Moles of hydrogen gas:

$\frac{11.0}{2}=5.5 moles$

2 moles of hydrogen reacts with 1 mole of oxygen then, 5.5 moles of hydrogen will react with 2.75 moles of oxygen gas.

Hydrogen gas is in limited amount hence will act as limiting reagent in the reaction.

According to reaction, 2 moles of hydrogen reacts gives 2 mole of water

Then, 5.5 moles of hydrogen gas will give:

$\frac{2}{2}\times 5.5 =5.5 moles$ of water

Theoretical yield of water = 5.5 × 18 g/mol=99 g

Percentage yiled:

$\frac{Experimental}}{Theoretical}\times 100=\frac{87 g}{99 g}\times 100=87.87\%$

The percent yield of water is 87.87%.