A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
\[{{M}_{2}}{{O}_{x}}\xrightarrow{\operatorname{Re}duction}M\] |
Eq. of \[{{M}_{2}}{{O}_{x}}=\] eq. of Metal |
\[\frac{Wt\,\,of\,\,{{M}_{2}}{{O}_{x}}}{Eq.\,wt.\,of\,{{M}_{2}}{{O}_{x}}}=\frac{Wt.\,of\,Metal}{Eq.\,wt.\,of\,Metal}\] |
\[\frac{4}{\frac{2\times 56+x\times 16}{2x}}=\frac{2.8}{\frac{56}{x}}\] |
On solving we get, |
\[\Rightarrow \,\,\frac{4}{56+8x}=\frac{2.8}{56}\Rightarrow \frac{1}{14+2x}=\frac{1}{20}\Rightarrow 2x=6\Rightarrow x=3\]Hence, the oxide is \[{{M}_{2}}{{O}_{3}}\]. |
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