\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,1 0,2
a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(C_{M_{ddHCl}}=\dfrac{0,2}{1,5}=0,13\left(M\right)\)
b) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,2 0,2
\(n_{NaOH}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(m_{NaOH}=0,2.40=8\left(g\right)\)
\(m_{ddNaOH}=\dfrac{8.100}{5}=160\left(g\right)\)
Chúc bạn học tốt
\(a,n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow n_{HCl}=2n_{Mg}=0,2\left(mol\right)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,2}{1,5}=\dfrac{2}{15}M\\ b,n_{HCl}=\dfrac{2}{15}\cdot0,75=0,1\left(mol\right)\\ PTHH:HCl+NaOH\rightarrow NaCl+H_2O\\ \Rightarrow n_{NaOH}=n_{HCl}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{NaOH}}=0,1\cdot40=4\left(g\right)\\ \Rightarrow m_{dd_{NaOH}}=\dfrac{4\cdot100\%}{5\%}=80\left(g\right)\)
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{HCl}=2.n_{Mg}=2.0,1=0,2\left(mol\right)\\ a,C_{M\text{dd}HCl}=\dfrac{0,2}{1,5}=\dfrac{2}{15}\left(M\right)\\ b,n_{HCl}=\dfrac{0,75}{1,5}.0,2=0,1\left(mol\right)\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{HCl}=0,1\left(mol\right)\\ m_{\text{dd}NaOH}=\dfrac{0,1.40.100}{5}=80\left(g\right)\)
