\(n_{Na_2CO_3}=\dfrac{m}{M}=\dfrac{5,3}{106}=0,05mol\)

\(n_{N_2CO_3}=\dfrac{m}{M}=\dfrac{5.3}{106}=0.05mol\)

Giup mk với

Ta có: \(m_{NaOH}=\dfrac{8\%.400}{100\%}=32\left(g\right)\)

\(m_{Na_2CO_3}=\dfrac{5,3\%.400}{100\%}=21,2\left(g\right)\)

\(\Rightarrow n_{NaOH}=\dfrac{32}{40}=0,8\left(mol\right)\)

\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

\(m_{Na_2CO_3}=\dfrac{100.5,3\%}{100\%}=5,3\left(g\right)\)

=> \(n_{Na_2CO_3}=\dfrac{5,3}{106}=0,05\left(mol\right)\)

Vậy số mol của 100g dd Na2CO3 5,3% là 0,05.

a) 

\(n_{SO_2}=\dfrac{m}{M}=\dfrac{3,2}{64}=0,05\left(mol\right)\\ n_{CO_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

b) 

\(n_{Cl_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\Rightarrow m_{Cl_2}=n.M=0,06.71=4,26\left(mol\right)\\ n_{Na_2CO_3}=n.M=0,5.106=53\left(g\right)\)

c) 

\(V_{N_2\left(\text{đ}ktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\\ n_{O_2}=\dfrac{m}{M}=\dfrac{4,8}{32}=0,15\left(mol\right)\Rightarrow V_{O_2\left(\text{đ}ktc\right)}=n.22,4=0,15.22,4=3,36\left(l\right)\)

Đáp án D

nNa2CO3 = \(\frac{700.5,3\%}{106100\%}=0,35mol\)

\(n_{CO2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_{NaOH}=0,6.2=0,6\left(mol\right)\)

Xét tỉ lệ :

\(\frac{n_{NaOH}}{n_{CO2}}=\frac{0,6}{0,25}=2,4>2\)

Tạo muối Na2CO3 và NaOH dư

\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

\(n_{Na2CO3}=n_{CO2}=0,25\left(mol\right)\)

\(\Rightarrow n_{NaOH\left(dư\right)}=0,6-0,25.2=0,1\left(mol\right)\)

Vậy đáp án đúng : B

\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

=> \(C_M=\dfrac{0,2}{0,3}=0,667M\)

\(m_{dd}=300.1,05=315\left(g\right)\)

=> \(C\%=\dfrac{21,2}{315}.100\%=6,73\%\)