Ừm , bài này mình đã làm ở dưới , bạn xem lại nha
Ta có: \(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\)
Ta lại có: \(C_{\%_{HCl}}=\dfrac{m_{HCl}}{91,25}.100\%=20\%\)
=> mHCl = 18,25 (g)
=> \(n_{HCl}=\dfrac{18,25}{36,5}=0,2\left(mol\right)\)
PTHH: CaCO3 + 2HCl ---> CaCl2 + H2O + CO2↑
Ta thấy: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\)
Vậy CaCO3 dư
Theo PT: \(n_{CO_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
=> \(V_{CO_2}=0,1.22,4=2,24\left(lít\right)\)
Theo PT: \(n_{H_2O}=n_{CaCl_2}=n_{H_2O}=0,1\left(mol\right)\)
=> \(m_{CaCl_2}=0,1.111=11,1\left(g\right)\)
\(m_{H_2O}=0,1.18=1,8\left(g\right)\)
=> \(m_{dd_{CaCl_2}}=20+91,25-1,8-\left(44.0,1\right)=105,05\left(g\right)\)
=> \(C_{\%_{CaCl_2}}=\dfrac{11,1}{105,05}.100\%=10,57\%\)
Ta có: \(m_{dd_{H_2O}}=20+91,25-11,1-\left(44.0,1\right)=95,75\left(g\right)\)
=> \(C_{\%_{H_2O}}=\dfrac{1,8}{95,75}.100\%=1,88\%\)
