$a) 2NaHCO_3 + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O + 2CO_2$

$b) n_{NaHCO_3}=\frac{21}{84}=0,25 (mol)$

$\text{Theo ptpư: }n_{H_2SO_4}=\frac{1}{2}n_{NaHCO_3}=0,125 (mol)$

$\Rightarrow V_{H_2SO_4}=\frac{0,125}{0,1}=1,25 (l)$

$c) \text{Theo ptpư: }n_{CO_2}=n_{NaHCO_3}=0,25 (mol)$

$\Rightarrow V_{CO_2}=0,25.22,4=5,6 (l)$

$m_{dd H_2SO_4}=1250.1,2=1500 (g)$

$m_{\text{dd sau}}=m_{NaHCO_3}+m_{dd \hspace{0,1cm}H_2SO_4}-m_{CO_2}$

$=21+1500-0,25.44=1510 (g)$

$m_{Na_2SO_4}=0,125.142=17,75 (g)$

$\Rightarrow C\%_{Na_2SO_4}=\frac{17,75}{1510}.100=1,176\%$

Đáp án:

 a/2NaHCO3 + H2SO4 --> Na2SO4 + 2CO2 + 2H2O

0,25               0,125            0,125         0,25

nNaHCO3=$\frac{21}{23+1+12+16.3}$ =0,25

b/VddH2SO4=$\frac{0,125}{0,1}$ =1,25lit

c/VCO2=0,25.22,4=5,6lit

1,25lit=1250ml

mddH2SO4=1250.1,2=1500g

mdd sau phản ứng=1500+21-0,25.44=1510

C%Na2SO4=$\frac{0,125.(23.2+32+16.4)}{1510}$ .100=1,18%

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