Calculate the amount of H 2 which is unreacted in the given reaction 2 H 2 + O 2 → 2 H 2 O If 8 g of H 2 is react with 16 g O 2 ?

Question

Calculate the amount of H 2 ​ which is unreacted in the given reaction 2 H 2 ​ + O 2 ​ → 2 H 2 ​ O If 8 g of H 2 is react with 16 g O 2 ?

TestHub · Accepted Answer

Explain the question : Find the remaining mass of $H_{2}$ after reaction of $8 g H_{2}$ with $16 g O_{2}$ .

Concept : This question is based on $O_{2}$ is the limiting reactant (LR) in the 2:1 mole ratio ( $2 H_{2} + O_{2} \to 2 H_{2} O) . H_{2}$ is in excess.

Solution :

1. Initial moles: H₂: $8 g /2 g / m o l = 4 m o l$ . O₂: $16 g /32 g / m o l = 0.5 m o l$ .

2. H₂ consumed: $0.5 m o l O_{2} t im es (2 m o l H_{2} /1 m o l O_{2}) = 1 m o l H_{2}$ .

3. H₂ unreacted: $(4 m o l - 1 m o l) t im es 2 g / m o l = 3 m o l t im es 2 g / m o l = 6 g$ .

Final answer: B

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