A) 2.16 g
B) 2.48 g
C) 2.64 g
D) 2.32 g
Correct Answer: A
Solution :
\[2A{{g}_{2}}C{{O}_{3}}\xrightarrow{\Delta }4Ag+2C{{O}_{2}}+{{O}_{2}}\] \[2\times 276\,gm\] \[4\times 108\,gm\] \[\because \] \[2\times 276gm\] of \[A{{g}_{2}}C{{O}_{3}}\] gives \[4\times 108\,gm\] \[\therefore \] \[1\,gm\] of \[A{{g}_{2}}C{{O}_{3}}\] gives \[=\frac{4\times 108}{2\times 276}\] \[\therefore \] \[2.76\,gm\] of \[A{{g}_{2}}C{{O}_{3}}\] gives \[\frac{4\times 108\times 2.76}{2\times 276}=2.16\,gm\]
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