A) 50%
B) 75%
C) 90%
D) 40%
Correct Answer: B
Solution :
\[Fe+2HC1\text{ }\left( excess \right)\xrightarrow{{}}FeC{{1}_{2}}+{{H}_{2}}\] 75 g Fe Vol. of \[{{H}_{2}}=area\times displacement\] \[=0.025\times 1\text{ }{{m}^{3}}=0.025\times {{10}^{3}}\text{ }lit=25\text{ }lit\] If Fe is x% pure \[E{{q}_{Fe}}=E{{q}_{{{H}_{2}}}}\] at 300 K; PV = nRT \[1\times 25=n\times 0.08\times 300\] \[n=1.04\,\,mol={{2.08}_{g}}{{H}_{2}}\] \[E{{q}_{Fe}}=E{{q}_{{{H}_{2}}}},\] \[\frac{0.75x}{\left( \frac{56}{2} \right)}=\frac{2.08}{1}\] x = 75% approx.You need to login to perform this action.
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