What will be the volume of CO_2 at S.T.P. obt | vedclass

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

Question

(A) The chemical equation for the decomposition of $BaCO_3$ is:$BaCO_3(s) 	o BaO(s) + CO_2(g)$Calculate the molar mass of $BaCO_3$:$M(BaCO_3) = 137 +

Vedclass · Accepted Answer

$1.12$

Vedclass · Answer

(A) The chemical equation for the decomposition of $BaCO_3$ is:$BaCO_3(s) 	o BaO(s) + CO_2(g)$Calculate the molar mass of $BaCO_3$:$M(BaCO_3) = 137 + 12 + (3 	imes 16) = 137 + 12 + 48 = 197 \ g/mol$According to the stoichiometry of the reaction,$1 \ mol$ of $BaCO_3$ produces $1 \ mol$ of $CO_2$.At $S.T.P.$,$1 \ mol$ of any gas occupies $22.4 \ L$.Therefore,$197 \ g$ of $BaCO_3$ produces $22.4 \ L$ of $CO_2$ at $S.T.P.$For $9.85 \ g$ of $BaCO_3$,the volume of $CO_2$ produced is:$V = \frac{22.4 \ L}{197 \ g} 	imes 9.85 \ g$$V = \frac{22.4 	imes 9.85}{197} = 1.12 \ L$Thus,the correct option is $A$.

What will be the volume of $CO_2$ at $S.T.P.$ obtained by heating $9.85 \ g$ of $BaCO_3$? (Atomic mass of $Ba = 137$)

Similar Questions

What volume of $0.1 \, M \, H_2SO_4$ is required to completely neutralize $40 \, mL$ of $0.2 \, M \, NaOH$?

$20 \ g$ of hydrogen gas $(H_2)$ is present in a $5 \ L$ vessel. The molar concentration of hydrogen is: (in $M$)

How much $g$ of $NaOH$ is required to neutralize $1500 \, cm^3$ of $0.1 \, N$ $HCl$ (At. wt. of $Na = 23$)?

If the density of liquid $HCl$ is $1.17 \, g \, cm^{-3}$,then its molarity will be ...............

The number of molecules of $CO_2$ liberated by the complete combustion of $0.1 \ mol$ of graphite in air is

Vedclass Test Series

Exam Paper Generator

Online Exam Module