X g of Ca [atomic mass = 40] dissolves compl | vedclass

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

Question

(C) The balanced chemical equation for the reaction is:$Ca + 2HCl \rightarrow CaCl_2 + H_2$At $STP$,$1 \ mole$ of any gas occupies $22.4 \ L$.Therefor

Vedclass · Accepted Answer

$9.0$

Vedclass · Answer

(C) The balanced chemical equation for the reaction is:$Ca + 2HCl ightarrow CaCl_2 + H_2$At $STP$,$1 \ mole$ of any gas occupies $22.4 \ L$.Therefore,the number of moles of $H_2$ produced is:$n(H_2) = \frac{5.04 \ L}{22.4 \ L/mol} = 0.225 \ mol$From the stoichiometry of the reaction,$1 \ mole$ of $Ca$ produces $1 \ mole$ of $H_2$.Thus,the moles of $Ca$ required is $0.225 \ mol$.Given the atomic mass of $Ca = 40 \ g/mol$,the mass $X$ is:$X = n(Ca) 	imes 	ext{molar mass}(Ca)$$X = 0.225 \ mol 	imes 40 \ g/mol = 9.0 \ g$Therefore,the value of $X$ is $9.0$.

$X \ g$ of $Ca$ [atomic mass $= 40$] dissolves completely in concentrated $HCl$ solution to produce $5.04 \ L$ of $H_2$ gas at $STP$. The value of $X$ is closest to

Similar Questions

Hardness of water is $200 \ ppm$. Calculate the molarity and normality of $CaCO_3$ in the water.

The volume in $mL$ of $0.1 \ M$ solution of $NaOH$ required to completely neutralize $100 \ mL$ of $0.3 \ M$ solution of $H_3PO_3$ is

At $T \ K$,$100 \ L$ of dry oxygen is present in a sealed container. It is subjected to silent electric discharge,until the volumes of oxygen and ozone become equal. What is the volume (in litres) of ozone formed at $T \ K$?

What is the ratio of the active masses of $22 \ g$ $CO_2$,$3 \ g$ $H_2$,and $7 \ g$ $N_2$ in a gaseous mixture?

$1 \ mol$ of metal '$M$' reacts completely with alcohol to give $1.5 \ mol$ of $H_2$. What is the valency of metal '$M$'?

Vedclass Test Series

Exam Paper Generator

Online Exam Module