A) 1008 g
B) 108 g
C) 1800 g
D) 3600 g
Correct Answer: A
Solution :
[a] This neutralization reaction will be : |
\[6CaO+{{P}_{4}}{{O}_{10}}\to 2C{{a}_{3}}{{(P{{O}_{4}})}_{2}}\] |
\[852g{{P}_{4}}{{O}_{10}}\equiv 3mol{{P}_{4}}{{O}_{10}}\] |
1 mole of \[{{P}_{4}}{{O}_{10}}\] neutralises 6 mol \[CaO\] |
3 moles of \[{{P}_{4}}{{O}_{10}}\]will netralise 18 moles of \[CaO\] |
Mass of \[CaO=18\times 56=1008g\] |
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