A) 90 kg
B) 540 kg
C) 270 kg
D) 180 kg
Correct Answer: A
Solution :
\[2A{{l}_{2}}{{O}_{3}}+3C\xrightarrow{{}}Al+3C{{O}_{2}}\] Gram equivalent of\[A{{l}_{2}}{{O}_{3}}\equiv gm\] equivalent of C. Now equivalent weight of \[Al=\frac{27}{3}=9\] Equivalent weight of \[C=\frac{12}{4}=3(\overset{0}{\mathop{C}}\,\to \overset{+4}{\mathop{C}}\,{{O}_{2}})\] No. of gram equivalent of Al \[Al=\frac{270\times {{10}^{3}}}{9}=30\times {{10}^{3}}\] Hence, No. of gram equivalent of\[C=30\times {{10}^{3}}\] Again, No. of gram equivalent of C \[=\frac{\text{mass}\,\text{in}\,\text{gram}}{\text{gram}\,\text{equivalent}\,\text{Weight}}\] \[\Rightarrow \]\[30\times {{10}^{3}}=\frac{mass}{3}\] \[\Rightarrow \]mass \[=90\times {{10}^{3}}\,g\,=90\,kg\]
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