Assuming fully decomposed, the volume of $$C{O_2}$$ released at $$STP$$ on heating $$9.85$$ $$g$$ of $$BaC{O_3}$$ ( Atomic mass, $$Ba = 137$$ ) will be
A.
1.12$$\,L$$
B.
2.24$$\,L$$
C.
4.06$$\,L$$
D.
0.84$$\,L$$
Answer :
1.12$$\,L$$
Solution :
$$BaC{O_3} \to BaO + C{O_2}$$
$$192\,g$$ of $$BaC{O_3}$$ gives $$1$$ $$mole$$ of $$C{O_2} = 22.4\,L$$
$$9.85\,g$$ of $$BaC{O_3}$$ will give $$0.05$$ $$mol$$ of $$C{O_2}$$ which is equal to $$1.12$$ $$litre.$$
Releted MCQ Question on Physical Chemistry >> Some Basic Concepts in Chemistry
Releted Question 1
$$27 g$$ of $$Al$$ will react completely with how many grams of oxygen?