How many mL of 22.4 volume H2O2 is required to oxidise 0.1 mol of H2S gas to S?

Question

How many mL of 22.4 volume H2O2 is required to oxidise 0.1 mol of H2S gas to S? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

underset1 textmolH2 O2+ underset1 textmolH2 S arrow 2H2O+S 0.1 mol H2S require H2O2=0.1 mol Molarity of 22.4 volume H2O2=(22.4/11.2)=2 textM Now volume of H2O2 required =(m o l/m o l a r i t y)=(0.1/2)L = (0.1 × 1000/2) = 50 textmL

Q. How many mL of 22.4 volume $H_{2} O_{2}$ is required to oxidise 0.1 mol of $H_{2} S$ gas to S?

$1 mol H_{2} O_{2} + 1 mol H_{2} S \to 2 H_{2} O + S$

0.1 mol $H_{2} S$ require $H_{2} O_{2} = 0.1$ mol

Molarity of 22.4 volume $H_{2} O_{2} = \frac{22.4}{11.2} = 2 M$

Now volume of $H_{2} O_{2}$ required

$= \frac{m o l}{m o l a r i t y} = \frac{0.1}{2} L$

$= \frac{0.1 \times 1000}{2} = 50 mL$

Q. How many mL of 22.4 volume H2​O2​ is required to oxidise 0.1 mol of H2​S gas to S?

1molH2​O2​​+1molH2​S​→2H2​O+S 0.1 mol H2​S require H2​O2​=0.1 mol Molarity of 22.4 volume H2​O2​=11.222.4​=2M Now volume of H2​O2​ required =molaritymol​=20.1​L =20.1×1000​=50mL

Q. How many mL of 22.4 volume $H_{2} O_{2}$ is required to oxidise 0.1 mol of $H_{2} S$ gas to S?

$1 mol H_{2} O_{2} + 1 mol H_{2} S \to 2 H_{2} O + S$

0.1 mol $H_{2} S$ require $H_{2} O_{2} = 0.1$ mol

Molarity of 22.4 volume $H_{2} O_{2} = \frac{22.4}{11.2} = 2 M$

Now volume of $H_{2} O_{2}$ required

$= \frac{m o l}{m o l a r i t y} = \frac{0.1}{2} L$

$= \frac{0.1 \times 1000}{2} = 50 mL$