Question

The answer mentioned for this is option 3 but I think the answer is option 2
What do you think? ​

Answer 1

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } GiveN:-}}}}}$

• 9.85 g $\sf \:BaCO_3$ is decomposed .
• Molecular weight of $\sf BaCO_3$ is 197 .

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } To\:FinD:-}}}}}$

• The Volume of $\sf CO_2$ obtained by its complete decomposition.

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } AnsweR:-}}}}}$

Given that 9.85g of Barium carbonate is decomposed . We know , when we decompose it Carbon dioxide and Barium oxide .

$\green{\tt \orange{\mapsto}The\: reaction\: would\:be:-}$

$\boxed{\red{\sf {\underset{\blue {Barium\: carbonate}}{BaCO_3}\xrightarrow{\green{Decomposition}}\underset{\blue{Barium\:oxide}}{BaO}+\underset{\blue{Carbon\: dioxide}}{CO_2}}}}$

This is a balanced chemical reaction and from the reaction it is clear that one mole of barium carbonate decomposes to give one mole of carbon dioxide.

$\sf{\purple\longrightarrow Let's\: calculate\:the\:number\:of\; moles:-}$

We can find number of moles as ,

$\boxed{\pink{\dag}\green{\bf n(moles)\:\:=\:\:\dfrac{Given\:Weight}{Molecular\:weight}}}$

$\tt Here$

• Given mass is 9.85g
• Gram Molecular mass is 197 g.

$\tt:\implies n(moles)=\dfrac{Given\:weight}{Molecular\: weight}$

$\tt:\implies n(moles)=\dfrac{9.85g}{197g}$

$\tt:\implies n(moles)=\dfrac{\cancel{985}}{100\times\cancel{197}}$

$\underline{\boxed{\red{\sf{\purple{\longmapsto}\:\:n(moles)\:\:\:\:=\:\:\:\:0.05}}}}$

Since one more of barium carbonate gives one mole of $\sf CO_2$hence the number of moles of $\sf CO_2$ formed will be 0.05.

$\sf \blue{Hence\:volume\:of\:CO_2\:at\: STP\:will\:be:-}$

$\tt:\implies Vol(CO_2)= V(STP) \times n(moles)$

$\tt:\implies Vol(CO_2)=22.4\times 0.05$

$\tt:\implies Vol(CO_2)=\dfrac{224}{10}\times\dfrac{5}{100}$

$\underline{\boxed{\red{\sf{\purple{\longmapsto}\:\:Vol(Carbon\: dioxide)\:\:\:\:=\:\:\:\:1 .12L}}}}$

$\boxed{\bf{\red{\dag}\blue{Hence\: the\: correct\:option\:is\:[b]\:1.12L.}}}$