A) 1.11 kg
B) 2.22kg
C) 3.33 kg
D) 4.44 kg
Correct Answer: D
Solution :
In 100 L of water amount of \[Ca(HC{{O}_{3}})=16.2\,g\] In 60,000 L of water amount of \[Ca{{(HC{{O}_{3}})}_{2}}=\frac{16.2\times 60,000}{100}\] \[=9720\,g\] \[=\frac{9720}{162}=60\,mol\] \[\underset{1\,mole}{\mathop{Ca{{(OH)}_{2}}}}\,+\underset{1\,mole}{\mathop{Ca{{(HC{{O}_{3}})}_{2}}}}\,\xrightarrow{{}}2CaC{{O}_{3}}+2{{H}_{2}}O\] Hence, amount of\[Ca{{(OH)}_{2}}\]required = 60 mot \[=60\times 74=4440\,g=4.44\text{ }kg\]
You need to login to perform this action.
You will be redirected in
3 sec
