\(n_{Mg}=\dfrac{7.2}{24}=0.3\left(mol\right)\)
\(n_{HCl}=\dfrac{21.9}{36.5}=0.6\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(0.3.......0.6......................0.3\)
\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(CuO+H_2\underrightarrow{^{^{t^o}}}Cu+H_2O\)
\(........0.3.....0.3\)
\(m_{Cu}=0.3\cdot64=19.2\left(g\right)\)
a) Mg + 2HCl \(\rightarrow\) MgCl2+ H2
b)nMg = \(\dfrac{7,2}{24}\)= 0,3(mol)
nHCl = \(\dfrac{21,9}{36,5}\)= 0,6(mol)
Mg + 2HCl \(\rightarrow\) MgCl2+ H2
\(\dfrac{n_{Mg}}{n_{2HCl}}=\dfrac{0,3}{0,6}\) \(\Rightarrow\)\(\dfrac{n_{Mg}}{n_{HCl}}=\dfrac{0,6}{0,6}\)=1
Mg + 2HCl \(\rightarrow\) MgCl2+ H2
(mol)03 --->0,6---------------->0,3
\(V_{H_2}\)= 0,3 .22,4 = 6,72(l)
c)
H2 + CuO \(\underrightarrow{t^o}\)Cu + H2O
(mol) 0,3------------>0,3
mCu = 0,3 . 64 = 19,2(g)
