\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a. Theo PT ta có: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b. Theo PT ta có: \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)
Mg + 2HCl → MgCl2 + H2
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
a) Theo pT: \(n_{H_2}=n_{Mg}=0,1\times22,4=2,24\left(l\right)\)
b) Theo pT: \(n_{HCl}=2n_{Mg}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,2\times36,5=7,3\left(g\right)\)
PTHH: Mg + 2HCl \(\rightarrow\) MgCl2 + H2\(\uparrow\)
a) nMg = \(\dfrac{2,4}{24}\)=0,1 (mol)
Theo PT: n\(H_2\) = nMg = 0,1 (mol)
\(\Rightarrow\)V H2 = 0,1 .22,4 = 2,24(l)
b) Theo PT : nHCl = 2nMg = 0,1.2 = 0,2 (mol)
\(\Rightarrow\) mHCl = 0,2.36,5 = 7,3(g)
PTHH : Mg + 2HCl → ZnCl2 + H2
a)\(n_{Mg}=\dfrac{V}{22,4}=\dfrac{2,4}{22,4}=\dfrac{3}{28}\left(mol\right)\\ PT=>n_{H_2}=n_{Mg}=\dfrac{3}{28}\left(mol\right)\\ ThểtíchkhíH_2là:\\ V_{H_2}=n_{H_2}.22,4=\dfrac{3}{28}.22,4=2,4\left(l\right)\\ \\ \\ \\ \\ \\ \\ b,PT=>n_{HCl}=\dfrac{2}{1}n_{Mg}=2.\dfrac{3}{28}=\dfrac{3}{14}\left(mol\right)\\ m_{HCl}=n_{HCl}.M_{HCl}=\dfrac{3}{14}.36,5=7,82\left(g\right)\)
