1)

m_{HCl}=200.36\%=72(g)

n_{HCl}={72}/{36,5}\approx 1,97(mol)

2)

V_{O_3(đkc)}=1,2.24,79=29,748(L)

3)

6197,5mL=6,1975L

n_{Cl_2}={6,1975}/{24,79}=0,25(mol)

m_{Cl_2}=0,25.71=17,75(g)

4)

50mL=0,05L

n_{FeCl_2}=0,05.3=0,15(mol)

m_{FeCl_2}=0,15.127=19,05(g)

5)

4958mL=4,958L

n_{N_2O}={4,958}/{24,79}=0,2(mol)

m_{N_2O}=0,2.44=8,8(g)

6)

n_{N_2}=7/{28}=0,25(mol)

V_{N_2(đkc)}=0,25.24,79=6,1975(L)

7)

m_{KOH}=125.32\%=40(g)

1)  n_(HCl)={m_(dd).C%}/(100.M)=(200.36)/(100.36,5)~~1,97(mol)

2)  V_{O_2(đkc)}=n.24,79=1,2.24,79=29,748(l)

3)  6197,5  ml = 6,1975 l

n_{Cl_2}={6,1975}/{24,79}=0,25(mol)

->  m_(Cl_2)=n.M=0,25.71=17,75(g)

4)  50  ml = 0,05l

n_{FeCl_2}=C_M.V=0,05.3=0,15(mol)

->  m_(FeCl_2)=n.M=0,15.127=19,05(g)

5)  4958  ml = 4,958l

n_{N_2O}=m/M={4,958}/{24,79}=0,2(mol)

->  m_{N_2O}=n.M=0,2.44=8,8(g)

6)  n_{N_2}=m/M=7/{28}=0,25(mol)

->  V_{N_2(đkc)}=0,25.24,79=6,1975(l)

7)  m_(KOH)={m_(dd).C%}/(100)=(125.32)/(100)=40(g)