Chemistry - Stoichiometry Question with Solution | TestHub

ChemistryStoichiometryCalculation of Percentage yieldMedium2 minQB

ChemistryMediummultiple choice

For the following reactions:

$(N H_{4})_{2} S O_{4} + N a O H 40% N a_{2} S O_{4} + H_{2} O + N H_{3}$

$N H_{3} + H Cl 80% N H_{4} Cl$

Choose correct option(s). [Assuming all other reactants required are in sufficient amount]

Answer:

A, B, C

Solution:

$n (N a O H) = \frac{4}{40} = 0.1 m o l$

Given a 40% yield, the amount of NH₃ produced is:

$0.1 \times 0.4 = 0.04 m o l$

With an 80% conversion, the amount of NH₄Cl formed is:

$0.04 \times 0.8 = 0.032 m o l$

(A)

For 10.7 g of NH₄Cl, the number of moles is:

$\frac{10.7}{53.5} = 0.2 m o l$

Considering the yield, the NaOH required is 25 g. (B, C)

Correct Answer: A, B, C

Stream:JEESubject:ChemistryTopic:StoichiometrySubtopic:Calculation of Percentage yield

⏱ 2mℹ️ Source: QB

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