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2H2 (1) + O2(g) → 2H20 (g)

1. Find the limiting reactant if you start with 30.0 grams of hydrogen and 5.29 grams of oxygen.

2. The actual yield for H2O in the above reaction is 6.72 g, Determine the percent yield for the reaction

when 9.93 grams of hydrogen and excess oxygen react?

User StanislavL
by
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1 Answer

10 votes

Answer: 1. Oxygen is the the limiting reactant.

2. 7.52%

Step-by-step explanation:

The balanced chemical equation is:


2H_2(g)+O_2(g)\rightarrow 2H_2O(g)

To calculate the moles :


\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}


\text{Moles of} H_2=(30.0g)/(2g/mol)=15.0moles


\text{Moles of} O_2=(5.29g)/(32g/mol)=0.165moles

According to stoichiometry :

1 mole of
O_2 require = 2 moles of
H_2

Thus 0.165 moles of
O_2 will require=
(2)/(1)* 0.165=0.331moles of
NH_3

Thus
O_2 is the limiting reagent as it limits the formation of product and
H_2 is the excess reagent.

2.
\text{Moles of} H_2=(9.93g)/(2g/mol)=4.96moles


\text{Moles of} H_2O=(6.72g)/(18g/mol)=0.373moles

As 2 moles of
H_2 give = 2 moles of
H_2O

Thus 4.96 moles of
H_2 give =
(2)/(2)* 4.96=4.96moles of
H_2O

percentage yield =
\frac{\text {actual yield}}{\text {theoretical yield}}=(0.373)/(4.96)* 100=7.52\%

Thus the percent yield for the reaction is 7.52%

User Samuel Zhang
by
6.3k points