CaOH2-CO2- CaCO3-H2O percentage yield - 30If 200 mL of N10CaOH2 is mixed with 2-2 g of CO2- then the number of moles of CaCO3 produced will be-

Given, Ca(OH)_2+CO_2⟶ CaCO_3+H_2O percentage yield=30 Number of moles of Ca(OH)_2 =Normality×1n_f× V(L) =110×12×2001000=0.01 mol Number of moles of CO_2 =Mass ...

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Byju's Answer

Standard XII

Chemistry

Redox Reactions

CaOH2+CO2⟶ Ca...

Question

$C a (O H)_{2} + C O_{2} ⟶ C a C O_{3} + H_{2} O percentage yield = 30%$
If $200 mL$ of $\frac{N}{10}$ $C a (O H)_{2}$ is mixed with $2.2 g$ of $C O_{2}$ , then the number of moles of $C a C O_{3}$ produced will be:

0.001

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0.003

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0.03

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0.01

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Solution

The correct option is B $0.003$
Given,
$C a (O H)_{2} + C O_{2} ⟶ C a C O_{3} + H_{2} O percentage yield = 30$

Number of moles of $C a (O H)_{2}$
$= Normality \times \frac{1}{n_{f}} \times V (L)$
$= \frac{1}{10} \times \frac{1}{2} \times \frac{200}{1000} = 0.01 mol$
Number of moles of $C O_{2}$
$= \frac{Mass}{Molar mass} = \frac{2.2}{44} = 0.05 mol$
Comparing the ratios of number of moles to the coefficient of each reactant and finding the smallest ratio will give us the limiting reagent.
$\frac{n_{C a (O H)_{2}}}{1} = 0.01$
$\frac{n_{C O_{2}}}{1} = 0.05$
$C a (O H)_{2}$ is the limiting reagent. So, the theoretical number of moles of $C a C O_{3}$ produced:
$\frac{n_{C a (O H)_{2}}}{1} = \frac{n_{C a C O_{3}}}{1}$
$∴ n_{C a C O_{3}} = 0.01 mol$
We know that:
$percentage yield = \frac{Actual yield}{theoretical yield} \times 100$

Actual number of mole of $C a C O_{3}$
$\frac{30}{100} \times 0.01 = 0.003 mol$

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Ca(OH)2+CO2⟶CaCO3+H2O percentage yield = 30% If 200 mL of N10 Ca(OH)2 is mixed with 2.2 g of CO2, then the number of moles of CaCO3 produced will be:

$C a (O H)_{2} + C O_{2} ⟶ C a C O_{3} + H_{2} O percentage yield = 30%$
If $200 mL$ of $\frac{N}{10}$ $C a (O H)_{2}$ is mixed with $2.2 g$ of $C O_{2}$ , then the number of moles of $C a C O_{3}$ produced will be: