Question

Answer 1

Câu 4:a) \(Na_2SO_3+2HCl\rightarrow2NaCl+SO_2+H_2O\)

 \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

Gọi \(n_{Na_2SO_3}=x\left(mol\right);n_{CaCO_3}=y\left(mol\right)\)

Theo đề ta có : \(\left\{{}\begin{matrix}126x+100y=69,1\\x+y=0,6\end{matrix}\right.\)

=> x=0,35 ; y=0,25

=> \(n_{NaCl}=2x=0,7\left(mol\right);n_{CaCl_2}=y=0,25\left(mol\right)\)

=> \(m_{NaCl}=40,95\left(g\right);m_{CaCl_2}=27,75\left(g\right)\)

b) \(m_{ddsaupu}=69,1+200-0,35.64-0,25.44=235,7\left(g\right)\)

=> \(C\%_{NaCl}=\dfrac{40,95}{235,7}.100=17,37\%\)

\(C\%_{CaCl_2}=\dfrac{27,75}{235,7}.100=11,77\left(g\right)\)

c) \(SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

\(n_{CaSO_3}=0,35\left(mol\right);n_{CaCO_3}=0,25\left(mol\right)\)

=> \(m_{kt}=0,35.120+0,25.100=67\left(g\right)\)

 

Answer 2

Câu 3 

a) Hiện tượng :Canxi nitrat phản ứng với Kali cacbonat tạo kết tủa trắng canxi cacbonat

PTHH: Ca(NO3)₂ + K₂CO₃ → CaCO₃ ↓ + 2KNO₃

b) \(n_{K_2CO_3}=\dfrac{69.20\%}{138}=0,1\left(mol\right)\)

Theo PT: \(n_{CaCO_3}=n_{K_2CO_3}=0,1\left(mol\right)\)

\(n_{CaCO_3}=n_{K_2CO_3}=0,1\left(mol\right)\)

=> \(m_{CaCO_3}=0,1.100=10\left(g\right)\)

\(V_{Ca\left(NO_3\right)_2}=\dfrac{0,1}{0,5}=0,2\left(lít\right)\)

c) \(CaCO_3-^{t^o}\rightarrow CaO+CO_2\)

\(n_{CaO}=n_{CaCO_3}=0,1\left(mol\right)\)

=> \(m_{CaO}=0,1.56=5,6\left(g\right)\)