Cân A
\(n_{Al}=\dfrac{20}{27}mol\\
2Al+6HCl\rightarrow2AlCl_3+3H_2\\
n_{H_2}=\dfrac{20}{27}\cdot\dfrac{3}{2}=\dfrac{10}{9}mol\\
m_{H_2}=m\downarrow=\dfrac{10}{9}=\dfrac{20}{9}\approx2,22g\)
Cân B
\(n_{Mg}=\dfrac{20}{24}=\dfrac{5}{6}mol\\
Mg+3HCl\rightarrow MgCl_2+H_2\\
n_{Mg}=n_{H_2}=\dfrac{5}{6}mol\\
m_{H_2}=m\downarrow=\dfrac{5}{6}\cdot2=\dfrac{5}{3}\approx1,67g\)
Vì \(m\downarrow_{cânA}>m\downarrow_{cânB}\) nên cân nghiêng về phía cân B
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