Question

Giúp mình với ạ

Answer 1

7.

\(n_{CO_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(n_{NaOH}=\dfrac{500\cdot1.3\cdot2.5\%}{40}=0.40652\left(mol\right)\)

\(T=\dfrac{0.40625}{0.5}=0.8125\)

=> Tạo muối axit

\(NaOH+CO_2\rightarrow NaHCO_3\)

\(0.40625.......................0.40625\)

\(C_{M_{NaHCO_3}}=\dfrac{0.40652}{0.5}=0.8125\left(M\right)\)

Answer 2

8.

\(n_{CO_2}=\dfrac{0.448}{22.4}=0.02\left(mol\right)\)

\(n_{NaOH}=0.1\cdot0.25=0.025\left(mol\right)\)

\(T=\dfrac{0.025}{0.02}=1.25\)

=> Tạo ra 2 muối 

\(n_{Na_2CO_3}=a\left(mol\right),n_{NaHCO_3}=b\left(mol\right)\)

Ta có : 

\(2a+b=0.025\)

\(a+b=0.02\)

\(\Rightarrow a=0.005,b=0.015\)

\(m_{Muối}=0.005\cdot106+0.015\cdot84=1.79\left(g\right)\)