11.
\(2NaCl+H_2SO_4\rightarrow Na_2SO_4+2HCl\)
\(n_{HCl}=\frac{50.14,6\%}{36,5}=0,2\left(mol\right)\)
\(\Rightarrow n_{NaCl}=0,2\left(mol\right)\)
\(\Rightarrow m_{NaCl}=0,2.58,5=11,7\left(g\right)\)
12.
mO trong oxit=\(3,33-2,13=1,2\left(g\right)\)
\(n_O=\frac{1,2}{16}=0,075\left(mol\right)\)
\(n_{HCl}=2n_O=0,075.2=0,15\left(mol\right)\)
\(\Rightarrow V_{HCl}=\frac{0,15}{2}=0,075\left(l\right)=75\left(ml\right)\)