Câu 1:
\(a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ \Rightarrow n_{MgCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,2\cdot95=19\left(g\right)\\ c,PTHH:H_2+CuO\rightarrow^{t^o}Cu+H_2O\\ \Rightarrow n_{Cu}=n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ \Rightarrow m_{Cu}=0,2\cdot64=12,8\left(g\right)\)
Câu 2:
\(a,PTHH:3Fe+2O_2\rightarrow^{t^o}Fe_3O_4\\ b,n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right);n_{Fe_3O_4}=\dfrac{13,92}{232}=0,06\left(mol\right)\\ \text{Vì }\dfrac{n_{Fe}}{3}< \dfrac{n_{Fe_3O_4}}{1}\Rightarrow Fe_3O_4\text{ dư}\\ \Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}\approx0,07\left(mol\right)\\ \Rightarrow V_{O_2}=0,07\cdot22,4=1,568\left(l\right)\)
