What is the volume of 0.1 , N , HCl required | vedclass

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

Question

(D) The chemical reaction is: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$. According to the law of equivalence,the number of equivalents of $HCl$

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(D) The chemical reaction is: $CaCO_3 + 2HCl ightarrow CaCl_2 + H_2O + CO_2$. According to the law of equivalence,the number of equivalents of $HCl$ must equal the number of equivalents of $CaCO_3$. Number of equivalents of $CaCO_3 = \frac{	ext{mass}}{	ext{equivalent weight}} = \frac{1.0 \, g}{50 \, g/eq} = 0.02 \, eq$. For $HCl$,the number of equivalents is given by $N 	imes V(L)$. $0.1 \, N 	imes V(L) = 0.02 \, eq$. $V(L) = \frac{0.02}{0.1} = 0.2 \, L$. Converting to $cm^3$: $0.2 \, L 	imes 1000 \, cm^3/L = 200 \, cm^3$.

What is the volume of $0.1 \, N \, HCl$ required to react completely with $1.0 \, g$ of pure calcium carbonate? (in $cm^3$)

Similar Questions

When $100 \, ml$ of $0.1 \, M$ $H_2SO_4$ reacts with $Na_2CO_3$, how many liters of $CO_2$ will be produced at $STP$ (in $L$)?

What will be the volume of $O_2$ gas at $STP$ for complete combustion of $7.8 \ g$ acetylene? ............ $L$

The density of a gas at $STP$ is $2.86 \, g \, L^{-1}$. Which of the following is the gas?

$A$ cylinder of gas is assumed to contain $11.2 \ kg$ of butane $(C_4H_{10})$. If a normal family needs $20000 \ kJ$ of energy per day,the cylinder will last: (Given that $\Delta H$ for combustion of butane is $-2658 \ kJ/mol$)....$days$

Calculate the amount of water $(g)$ produced by the combustion of $16 \ g$ of methane. (in $g$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module