A) 1.12L
B) 0.84L
C) 2.24 L
D) 4.06 L
Correct Answer: A
Solution :
\[\underset{\begin{smallmatrix} 137+42+48 \\ =197\,g \end{smallmatrix}}{\mathop{BaC{{O}_{3}}}}\,\xrightarrow{{}}BaO+\underset{22.4L\text{ }at\text{ }STP}{\mathop{C{{O}_{2}}}}\,\] \[\because \]197 g of\[BaC{{O}_{3}}\]gives\[22.4\text{ }L\text{ }C{{O}_{2}}\]at STP \[\therefore \]9.85 g of\[BaC{{O}_{3}}\]gives\[=\frac{22.4}{197}\times 9.85\] \[=1.12L\]You need to login to perform this action.
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