\(n_{Ca}=\dfrac{m}{M}=\dfrac{80}{40}=2\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{C\%\cdot mdd}{100\cdot M}=\dfrac{20\cdot140}{100\cdot98}=0,2857\left(mol\right)\)
PTHH : Ca + H2SO4 ----> CaSO4 + H2
............1...........1..................1..........1 ( tỉ lệ mol )
Theo PT : nCa = nH2SO4
Theo GT : nCa > nH2SO4 => Ca dư = > nCa dư = 2 - 0,2857 = 1,7143 ( mol)
PTHH ( Ca dư phản ứng với H2O )
Ca + 2H2O ---> Ca(OH)2 + H2
...1,7143...3,4286.......1,7143......1,7143... (mol)
=> \(V_{H_2SO_4}=\dfrac{140}{1,12}=125ml\)
mdd sau phản ứng = 80 + 140 = 220(g)
\(m_{CaSO_4}=n\cdot M=0,2857\cdot136=38,8552\left(g\right)\Rightarrow C\%=\dfrac{38,8552}{220}\cdot100\%=17,66\%\)
\(m_{Ca\left(OH\right)_2}=n\cdot M=0,7143\cdot74=52,8582\left(g\right)\Rightarrow C\%=\dfrac{52,8582}{220}\cdot100\%=24,03\%\)
