10 g of a piece of marble (CaCO_3) was put i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The chemical equation for the reaction is: $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$.According to the stoichiometry,$1 \ m

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(C) The chemical equation for the reaction is: $CaCO_3(s) + 2HCl(aq) ightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$.According to the stoichiometry,$1 \ mol$ of $CaCO_3$ $(100 \ g)$ produces $22400 \ mL$ of $CO_2$ at $STP$.If the marble were $100 \%$ pure,$10 \ g$ of $CaCO_3$ would produce: $\frac{22400 \ mL}{100 \ g} 	imes 10 \ g = 2240 \ mL$ of $CO_2$.However,only $1120 \ mL$ of $CO_2$ was obtained.The percentage purity of $CaCO_3$ is calculated as: $\frac{	ext{Actual volume of } CO_2}{	ext{Theoretical volume of } CO_2} 	imes 100 = \frac{1120 \ mL}{2240 \ mL} 	imes 100 = 50 \%$.

$10 \ g$ of a piece of marble $(CaCO_3)$ was put into excess of dilute $HCl$ acid. When the reaction was complete,$1120 \ mL$ of $CO_2$ was obtained at $STP$. The percentage of pure $CaCO_3$ in the marble is $............ \%$.

Similar Questions

When a hydrocarbon $A$ undergoes combustion in the presence of air,it requires $9.5$ equivalents of oxygen and produces $3$ equivalents of water. What is the molecular formula of $A$?

If $50 \ mL$ of $0.5 \ M$ oxalic acid is required to neutralize $25 \ mL$ of $NaOH$ solution, the amount of $NaOH$ in $50 \ mL$ of the given $NaOH$ solution is . . . . . . $g$.

The vapour density of a gas is $11.2$. The volume occupied by $11.2 \ g$ of this gas at $N.T.P.$ is ................ $L$

If $20 \ mL$ of $0.4 \ N \ NaOH$ solution completely neutralises $40 \ mL$ of a dibasic acid,the molarity of the acid solution is $.......... \ M$.

Find the mass of sodium carbonate in grams required to prepare $100 \text{ ml}$ of a $0.1 \text{ M}$ aqueous solution. (Molar mass of $Na_2CO_3 = 106 \text{ g/mol}$) (in $\text{ g}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module