92% pure Sodium carbonate is used in the reation Na₂CO₃ + CaCl₂→CaCO₃ + 2NaCl. The number of grams of Na₂CO₃ required to yield one gram of CaCO₃

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8.5g

10.5g

11.52g

1.152g

answer is D.

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Detailed Solution

$large mathop {mathop {mathop {N{a_2}C{O_3}}limits_{106g} }limits_{'x'g} }limits^{1;mole} + {text{ }}CaC{l_{2;}} to mathop {mathop {mathop {CaC{O_3}}limits_{100g} }limits_{1g} }limits^{1;mole} + {text{ }}2NaCl$
$large X; = ;frac{{1 times 106}}{{100}}; = ;1.06g$
But the purity of Na₂CO₃ is only 92% thus more than 1.06g of Na₂CO₃ is requied to produce 1gram of CaCO₃
$large therefore$ Wt of Na₂CO₃ required = $large frac{100}{92}times 1.06$
= 1.152g