\(a.m_{HCl}=\dfrac{150.12}{100}=18\left(g\right)\\ n_{HCl}=\dfrac{18}{36,5}\approx0,5\left(mol\right)\\ b.m_{H_2SO_4}=\dfrac{50.10}{100}=5\left(g\right)\\ n_{H_2SO_4}=\dfrac{5}{98}\approx0,05\left(mol\right)\\ c.m_{AgNO_3}=\dfrac{100.12}{100}=12\left(g\right)\\ n_{AgNO_2}=\dfrac{12}{170}\approx0,07\left(mol\right)\)
a.
\(m_{HCl}=\dfrac{150.12\%}{100\%}=18\left(g\right)\\ \Rightarrow n_{HCl}=\dfrac{18}{36,5}\approx0,5\left(mol\right)\)
b.
\(m_{H_2SO_4}=\dfrac{50.10\%}{100\%}=5\left(g\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{5}{98}\approx0,05\left(mol\right)\)
c.
\(m_{AgNO_3}=\dfrac{100.12\%}{100\%}=12\left(g\right)\\ \Rightarrow n_{AgNO_3}=\dfrac{12}{170}\approx0,07\left(mol\right)\)