\(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\\ a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{MgCl_2}=n_{H_2}=n_{Mg}=0,4\left(mol\right)\\ m_{MgCl_2}=95.0,4=38\left(g\right)\\ b,V_{H_2\left(đkc\right)}=0,4.24,79=9,916\left(l\right)\\ d,n_{HCl}=0,4.2=0,8\left(mol\right)\\ V_{ddHCl}=\dfrac{0,8}{2}=0,4\left(l\right)\)
Đúng 1
Bình luận (2)