Bài 4 :
a. PTHH : 2HgO -to-> 2Hg + O2
8 4
\(V_{O_2}=4.22,4=89,6\left(l\right)\)
b. \(n_{HgO}=\dfrac{434}{217}=2\left(mol\right)\)
PTHH : 2HgO -to-> 2Hg + O2
2 2
\(m_{Hg}=2.201=402\left(g\right)\)
c. \(n_{Hg}=\dfrac{150,75}{201}=0,75\left(mol\right)\)
PTHH : 2HgO -to-> 2Hg + O2
0,75 0,75
\(m_{HgO}=0,75.217=162,75\left(g\right)\)
Bài 5:
\(2Al+3Cl_2\rightarrow2AlCl_3\)
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(\Leftrightarrow n_{AlCl_3}=0.2\left(mol\right)\)
\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)
\(n_{Cl_2}=0.3\left(mol\right)\)
\(V=0.3\cdot22.4=6.72\left(lít\right)\)
Bài 5 :
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH : 2Al + 3Cl2 -> 2AlCl3
0,2 0,3 0,2
\(V_{Cl_2}=0,3.22,4=6,72\left(l\right)\)
\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
