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ICSE Class 10 Chemistry 2015 Paper

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Consider the following reaction and based on the reaction answer the questions that follow :
(NH4)2Cr2O7→heatN2(g)+4H2O(g)+Cr2O3(\mathrm{NH}_4)_2\mathrm{Cr}_2\mathrm{O}_7 \xrightarrow{\text{heat}} \mathrm{N}_2(g)+4\mathrm{H}_2\mathrm{O}(g)+\mathrm{Cr}_2\mathrm{O}_3
Calculate:
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Question : 60 of 85
Marks: +1, -0
the quantity in moles of (NH4)2Cr2O7\mathrm{(NH_4)_2Cr_2O_7} if 63 gm63\ \mathrm{gm} of (NH4)2Cr2O7\mathrm{(NH_4)_2Cr_2O_7} is heated.
Solution:  
(NH4)2Cr2O7→Δ  N2↑+4H2O+  Cr2O32(14+4)+(52×2)=2×14  =2×52+(16×7)=28 gm+3×16=36+104+112=252 gm=152 gm\begin{array}{lllllll} \mathrm{(NH_4)_2Cr_2O_7} & \xrightarrow{\Delta} & \;\mathrm{N_2}\uparrow & + & 4\mathrm{H_2O} & + & \;\mathrm{Cr_2O_3} \\ 2(14+4)+(52\times 2) & & =2\times 14\; & & & & =2\times 52 \\ +(16\times 7) & & =28\ \mathrm{gm} & & & & +3\times 16 \\ =36+104+112=252\ \mathrm{gm} & & & & & & =152\ \mathrm{gm} \end{array}
  No. of Moles  =    Wt. of  (NH4)2Cr2O7  Gram. Mol. Mass  \; \text{No. of Moles}\; = \; \frac{\; \text{Wt. of}\; \mathrm{(NH_4)_2Cr_2O_7}}{\; \text{Gram. Mol. Mass}\; }
  =  63252=0.25  moles  \; = \; \frac{63}{252} = 0.25 \; \text{moles}\;
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