150ml=0,15l
\(n_{H_2SO_4}=2.0,15=0,3mol\)
\(m_{H_2SO_4}=0,3.98=29,4g\)
\(n_{H_2SO_4}=V_{ddH_2SO_4}\cdot C_{M_{H_2SO_4}}=0,15\cdot2=0,3mol\)
\(\Rightarrow m_{H_2SO_4}=0,3\cdot98=29,4g\)
Đổi \(150ml=0,15l\)
\(n_{H_2SO_4}=C_M.V=2.0,15=0,3\left(mol\right)\)
\(m_{H_2SO_4}=n.M=0,3.98=29,4\left(g\right)\)