Mg + 2HCl \(\rightarrow\) MgCl2 + H2
a) nHCl = \(\frac{m}{M}\) = \(\frac{7,1}{36,5}\) = 0,2 mol
Theo PTHH, ta có:
nMg = \(\frac{0,2.1}{2}\) = 0,1 mol
mMg = n.M = 0,1.24 = 2,4 g
b) Theo PTHH, ta có:
n\(H_2\) = \(\frac{0,2.1}{2}\) = 0,1 mol
V\(H_2\) = n.22,4 = 0,1.22,4= 2,24l (đktc)
c) Theo PTHH, ta có:
n\(MgCl_2\) = \(\frac{0,2.1}{2}\) = 0,1 mol
m\(MgCl_2\) = n.M = 0,1.95 = 9,5 g
Mg + 2HCl --> MgCl2 + H2
nHCl = 7.1/36.5 \(\approx\)0.2 mol
Từ PTHH:
=> nMg = 0.1 mol
mMg = 2.4 g
=> nH2 = 0.1 mol
VH2 = 0.1*22.4=2.24 l
nMgCl2 = 0.1 mol
mMgCl2 = 0.1*95=9.5g
Mg + 2HCl → MgCl2 + H2
\(n_{HCl}=\frac{7,1}{36,5}=\frac{71}{365}\left(mol\right)\)
a) Theo PT: \(n_{Mg}=\frac{1}{2}n_{HCl}=\frac{1}{2}\times\frac{71}{365}=\frac{71}{730}\left(mol\right)\)
\(\Rightarrow m_{Mg}=\frac{71}{730}\times24=2,33\left(g\right)\)
b) Theo Pt: \(n_{H_2}=n_{Mg}=\frac{71}{730}\left(mol\right)\)
\(\Rightarrow V_{H_2}=\frac{71}{730}\times22,4=2,18\left(l\right)\)
c) Theo PT: \(n_{MgCl_2}=n_{Mg}=\frac{71}{730}\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=\frac{71}{730}\times95=9,24\left(g\right)\)