A) 75
B) 30.6
C) 25
D) 69.4
Correct Answer: A
Solution :
\[\underset{1\,\,mol}{\mathop{CaC{{O}_{3}}}}\,\xrightarrow{\Delta }\underset{1\,\,mol}{\mathop{CaO}}\,+C{{O}_{2}}\] \[\underset{1\,\,mol}{\mathop{CaC{{l}_{2}}}}\,+N{{a}_{2}}C{{O}_{3}}\xrightarrow{{}}\underset{1\,\,mol}{\mathop{CaC{{O}_{3}}}}\,+2Na\] \[1\,\,mol\,\,CaO\cong 1\,\,mol\,\,CaC{{l}_{2}}\] \[\frac{0.56}{56}mol\,\,CaO\cong 0.01\,\,mol\,\,CaC{{l}_{2}}\] \[=0.01\times 111\,\,g\,\,CaC{{l}_{2}}\] \[=1.11\,\,g\,CaC{{l}_{2}}\] Thus, in the mixture, weight of \[NaCl=4.44-1.11=3.33\,\,g\] \[\therefore \] Percentage of \[NaCl=\frac{3.33}{4.44}\times 100\] = 75 %You need to login to perform this action.
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