0.500 g of a copper alloy is dissolved and treated with excess KI (2Cu^2+ + 4I^- -> 2CuI + I2). The iodine liberated needs 27.4 mL of 0.100 M Na2S2O3. Taking Cu = 63.5, find the mass percent of copper (one decimal).

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$0.500 g$ of a copper alloy is dissolved and treated with excess $KI$ ( $2 Cu X^{2 +} + 4 I X^{-} 2 CuI + I X_{2}$ ). The iodine liberated needs $27.4 mL$ of $0.100 M$ $Na X_{2} S X_{2} O X_{3}$ . Taking Cu = $63.5$ , find the mass percent of copper (one decimal).

Numerical answer type, answer in %, tolerance 0.1.

Answer: 34.8 %

mol thiosulphate = 0.0274 x 0.100 = 2.74e-3; mol Cu2+ = same (1:1). Mass Cu = 2.74e-3 x 63.5 = 0.1740 g. Percent = 0.1740/0.500 x 100 = 34.80, about 34.8%.

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0.500 g of a copper alloy is dissolved and treated with excess KI (2CuX2++4IX−​2CuI+IX2​). The iodine liberated needs 27.4 mL of 0.100 M NaX2​SX2​OX3​. Taking Cu = 63.5, find the mass percent of copper (one decimal).

Answer: 34.8 %

More questions on Iodometric and iodimetric titrations

$0.500 g$ of a copper alloy is dissolved and treated with excess $KI$ ( $2 Cu X^{2 +} + 4 I X^{-} 2 CuI + I X_{2}$ ). The iodine liberated needs $27.4 mL$ of $0.100 M$ $Na X_{2} S X_{2} O X_{3}$ . Taking Cu = $63.5$ , find the mass percent of copper (one decimal).