A) 2.16 g
B) 2.48 g
C) 2.64 g
D) 2.32 g
Correct Answer: A
Solution :
\[2A{{g}_{2}}C{{O}_{3}}\xrightarrow{\Delta }4Ag+\underset{s}{\mathop{2C{{O}_{2}}}}\,+{{O}_{2}}\] \[2\times [108\times 2+12+3\times 16]\] \[2\times [216+60]\] \[2\times 276g\,\,\,\,4\times 108\,g\] \[\because \] \[2\times 276g\,of\,\,A{{g}_{2}}C{{O}_{3}}gives\] \[=4\times 108g\,Ag\] \[\therefore \] 1 g \[A{{g}_{2}}C{{O}_{3}}\]gives \[=\frac{4\times 108}{2\times 276}\] \[\therefore \] \[2.76\,\,g\,\,of\,A{{g}_{2}}C{{O}_{3}}\,\,\,gives\] \[=\frac{4\times 108\times 2.76}{2\times 276}\] \[=2.16\,g\]
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