x 196
2NaCl + H₂SO₄ --------> Na₂SO₄ + 2HCl
2*58.5 98
x = [(2*58.5)*196] / 98 = 234g NaCl
SAU
x = 4 moli 2 moli
2NaCl + H₂SO₄ ----------> Na₂SO₄ + 2HCl
2 moli 1 mol
[tex]n_{_{H_2SO_4}}=\frac{m}{M}=\frac{196\ {\not}g}{98\ {\not}g/mol}=2\ moli\\\\n_{NaCl}=x=4\ moli\\\\m_{NaCl}=n*M=4\ {\not}moli*58.5\ g/{\not}mol=234\ g[/tex]