During the preparation of H2S2O8 (per disulphuric acid) O2 gas is also released at anode as byproduct. When 9.72 L of H2 releases at cathode and 2.35 L O2 at anode, the weight H2S2O8 produced in gram is

Question

During the preparation of H2S2O8 (per disulphuric acid) O2 gas is also released at anode as byproduct. When 9.72 L of H2 releases at cathode and 2.35 L O2 at anode, the weight H2S2O8 produced in gram is (A) 87.12 (B) 43.65 (C) 83.42 (D) 51.74

Tardigrade Admin · Accepted Answer

2HSO-4 → H2S2O8 +2e 2O2- → O2 +4e (2.35/22.4)=0.105 0.42 mole 2H+ +2e → H2 0.87 mole (9.72/22.4) mole mole of e for H2S2O8 = 0.87 - 0.42 = 0.45 mole ∴ mole of H2S2O8 = 0.225 mole = 43.65 g

Q. During the preparation of $H_{2} S_{2} O_{8}$ (per disulphuric acid) $O_{2}$ gas is also released at anode as byproduct. When $9.72 L$ of $H_{2}$ releases at cathode and $2.35 L$ $O_{2}$ at anode, the weight $H_{2} S_{2} O_{8}$ produced in gram is

$87.12$

$43.65$

$83.42$

$51.74$

$2 H S O_{4}^{-} \to H_{2} S_{2} O_{8} + 2 e$
$2 O^{2 -} \to O_{2} + 4 e$
$\frac{2.35}{22.4} = 0.1050.42$ mole
$2 H^{+} + 2 e \to H_{2}$
$0.87$ mole $\frac{9.72}{22.4}$ mole
mole of e for $H_{2} S_{2} O_{8} = 0.87 - 0.42 = 0.45$ mole
$∴$ mole of $H_{2} S_{2} O_{8} = 0.225$ mole $= 43.65$ g

Q. During the preparation of H2​S2​O8​ (per disulphuric acid) O2​ gas is also released at anode as byproduct. When 9.72L of H2​ releases at cathode and 2.35L O2​ at anode, the weight H2​S2​O8​ produced in gram is

87.12

43.65

83.42

51.74