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Đáp án:

$\rm n_{NaCl}=\dfrac{12,87}{58,5}=0,22\ (mol)$

Phương trình phản ứng:

$\rm 2NaCl + H_2SO_4(conc .\!)\xrightarrow{t^°} Na_2SO_4 + 2HCl$

Theo phương trình:

$\rm n_{Na_2SO_4}=\dfrac{1}{2}n_{NaCl}=0,11\ (mol)\\n_{HCl}=n_{NaCl}=0,22\ (mol)$

$\rm\to m_{Na_2SO_4}=0,11×142×90\%=14,058\ (g)\\\to V_{HCl}=0,22×22,4×90\%=4,4352\ (L)$

$\rm n_{NaCl}=\dfrac{12,87}{58,5}=0,22(mol)$
$\rm PTHH:2NaCl+H_2SO_4 \xrightarrow{t^o>400^oC} Na_2SO_4+2HCl↑$

$\rm n_{Na_2SO_4}=\dfrac{1}{2}n_{NaCl}=0,11(mol)$ 

$\rm n_{HCl}=n_{NaCl}=0,22(Mol)$

$\rm H=90\%$
$\rm m_{Na_2SO_4}=0,11.142.90\%=14,058(g)$

$\rm V_{HCl}=0,22.22,4.90\%=4,4352(l)$