17.0 g of AgNO3 is mixed with 4.68 g of NaCl in water: AgNO3 + NaCl -> AgCl + NaNO3. If the isolated AgCl is 92% of the theoretical amount, find its mass in grams to two decimals (Ag = 108, N = 14, O = 16, Na = 23, Cl = 35.5).

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JEE Advanced levelNumerical answerDifficulty 3 of 3mc198

$17.0 g$ of $AgNO X_{3}$ is mixed with $4.68 g$ of $NaCl$ in water: $AgNO X_{3} + NaCl AgCl + NaNO X_{3}$ . If the isolated $AgCl$ is $92%$ of the theoretical amount, find its mass in grams to two decimals (Ag = 108, N = 14, O = 16, Na = 23, Cl = 35.5).

Numerical answer type, answer in g, tolerance 0.01.

Answer: 10.56 g

n(AgNO3) = 17.0/170 = 0.10 mol; n(NaCl) = 4.68/58.5 = 0.08 mol, so NaCl is limiting. Theoretical AgCl = 0.080 x 143.5 = 11.48 g. At 92% yield: 10.56 g.

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17.0 g of AgNOX3​ is mixed with 4.68 g of NaCl in water: AgNOX3​+NaCl​AgCl+NaNOX3​. If the isolated AgCl is 92% of the theoretical amount, find its mass in grams to two decimals (Ag = 108, N = 14, O = 16, Na = 23, Cl = 35.5).

Answer: 10.56 g

More questions on Limiting reagent, percent yield and percent excess

$17.0 g$ of $AgNO X_{3}$ is mixed with $4.68 g$ of $NaCl$ in water: $AgNO X_{3} + NaCl AgCl + NaNO X_{3}$ . If the isolated $AgCl$ is $92%$ of the theoretical amount, find its mass in grams to two decimals (Ag = 108, N = 14, O = 16, Na = 23, Cl = 35.5).