Answer:
Calculation of mass of \[C{{O}_{2}}\]
produced:
Mass of the compound
= 0.20 g
We know
that,
\[%\,of\,C=\frac{12}{44}\times
\frac{Mass\,of\,C{{O}_{2}}\,formed}{Mass\,of\,the\,compound}\times 100\]
\[69=\frac{12}{44}\times
\frac{Mass\,of\,C{{O}_{2}}\,formed}{0.2}\times 100\]
or mass of \[C{{O}_{2}}\]
formed =
\[\frac{69\times
44\times 0.2}{12\times 100}=0.56g\]
Similarly,
\[%\,of\,H=\frac{2}{18}\times
\frac{Mass\,of\,{{H}_{2}}O\,formed}{Mass\,of\,the\,compound}\times 100\]
\[4.8=\frac{2}{18}\times
\frac{Mass\,of\,{{H}_{2}}O\,formed}{0.2}\times 100\]
or mass of \[{{H}_{2}}O\]
formed =
\[=\frac{4.8\times
18\times 0.2}{2\times 100}=0.0864g\]
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