Test Index

ICSE Class 10 Chemistry 2016 Paper

© examsnet.com
Question : 27 of 85
Marks: +1, -0
How much Calcium oxide is formed when 82 g82\ \mathrm{g} of calcium nitrate is heated? Also find the volume of nitrogendioxide evolved :
  2 Ca(NO3)2→2 CaO+4 NO2+O2\;2\ \mathrm{Ca}(\mathrm{NO}_3)_2 \rightarrow 2\ \mathrm{CaO}+4\ \mathrm{NO}_2+\mathrm{O}_2
  (Ca=40, N=14, O=16)\;(\mathrm{Ca}=40,\ \mathrm{N}=14,\ \mathrm{O}=16)
Solution:  
2 Ca(NO3)2→Δ2 CaO+4 NO2+O2  2[40+2(14+48)]2(40+16)4   moles =328 gm=112 gm=4×22.4=89.6 l\begin{array}{lllllll} 2\ \mathrm{Ca}(\mathrm{NO}_3)_2 & \xrightarrow{\Delta} & 2\ \mathrm{CaO} & + & 4\ \mathrm{NO}_2 & + & \mathrm{O}_2 \\ \;2[40+2(14+48)] & & 2(40+16) & & & 4\;\text{ moles } & \\ =328\ \mathrm{gm} & & =112\ \mathrm{gm} & & & =4 \times 22.4 & \\ & & & & =89.6\ \mathrm{l} & & \end{array}
1. If 328 g328\ \mathrm{g} of Ca(NO3)2\mathrm{Ca}(\mathrm{NO}_3)_2 releases 112 gm\mathrm{gm} of CaO\mathrm{CaO}
Then 82 g82\ \mathrm{g} of Ca(NO3)2\mathrm{Ca}(\mathrm{NO}_3)_2 will    releases   =  112328×82\;\text{ releases }\;=\;\frac{112}{328} \times 82
=28 gms   of   CaO=28\ \mathrm{gms}\;\text{ of }\;\mathrm{CaO}
2. If 328 g328\ \mathrm{g} of Ca(NO3)2\mathrm{Ca}(\mathrm{NO}_3)_2 releases 89.6 l89.6\ \mathrm{l} of NO2\mathrm{NO}_2 at STP
Then 82 g82\ \mathrm{g} of Ca(NO3)2\mathrm{Ca}(\mathrm{NO}_3)_2 will releases
  =  89.6328×82\;=\;\frac{89.6}{328} \times 82
  =22.4   litres of   NO2.\;=22.4\;\text{ litres of }\;\mathrm{NO}_2.
© examsnet.com
Go to Question:
Prev Question
Next Question