Mg + 2HCl → \(MgCl_2\)+\(H_2\)
0,1 0,05 0,05 (mol)
a) V= 50ml = 0,05 l
\(n_{HCl}\)= \(C_M\). V= 2 . 0,05= 0,1 (mol)
\(V_{H_2}\)= n . 22,4 = 0,1 . 22,4 = 2,24 (l)
b) \(C_{M/ddsauphảnứng}\)= \(\dfrac{n}{V}\)= \(\dfrac{0,05}{0,05}\)= 1 (M)
Mg + 2HCl → MgCl2 + H2
\(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)
\(n_{HCl}=0,05\times2=0,1\left(mol\right)\)
Theo PT: \(n_{Mg}=\dfrac{1}{2}n_{HCl}\)
Theo bài: \(n_{Mg}=\dfrac{1}{2}n_{HCl}\)
Vì \(\dfrac{1}{2}=\dfrac{1}{2}\) ⇒ HCl và Mg đều hết
a) Theo PT: \(n_{H_2}=n_{Mg}=0,05\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,05\times22,4=1,12\left(l\right)\)
b) Theo PT: \(n_{MgCl_2}=n_{Mg}=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,05}{0,05}=1\left(M\right)\)
