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when 10g of $C a C O_{3}$ is treated with 10g of HCl, mass of $C O_{2}$ will produced is $n \times 10^{- 1} .$ Then the
value of n is

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answer is 44.

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Detailed Solution

$C a C O_{3} + 2 H C l \to C a C l_{2} + H_{2} O + C O_{2} 100 73 44$

$C a C O_{3}$ is limiting reagent

100g $C a C O_{3}$ → 44g of $C O_{2}$
10 g $C a C O_{3}$ →
mass of $C O_{2}$ produced = $\frac{440}{100}$ =4.4 g

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