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