TN
DL
23 tháng 8 lúc 8:09

1. \(n_{H_2}=n_{MgSO_4}=0,02\left(mol\right)\Rightarrow V_{H_2}=0,02.24,79=0,4958\left(l\right)\)

2. \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{5}{4}.n_P=0,25\left(mol\right)\Rightarrow V_{O_2}=0,25.24,79=6,1975\left(l\right)\\ n_{P_2O_5}=\dfrac{2}{4}.n_P=0,1\left(mol\right)\Rightarrow m_{P_2O_5}0,1.142=14,2\left(g\right)\)

3. \(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)

  \(2Al+\dfrac{3}{2}O_2\underrightarrow{t^o}Al_2O_3\)

0,02->0,015--->0,01

\(m_{Al_2O_3}=0,01.102=1,02\left(g\right)\)

\(V_{O_2}=0,015.24,79=0,37185\left(l\right)\)

4. \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1-------------> 0,1-----> 0,1

\(m_{FeCl_2}=0,1.127=12,7\left(g\right)\\ V_{H_2}=0,1.24,79=2,479\left(l\right)\)

Bình luận (0)