3SnCl₂ + K₂Cr₂O₇ <span>+ 14HCl = 3</span>SnCl₄ + 2CrCl₃ <span>+ 2KCl + 7H</span>₂<span>O
</span>Sn⁺²Cl₂⁻ + K₂⁺Cr₂⁺⁷O₇²⁻ + H⁺Cl⁻ = Sn⁺⁴Cl₄⁻ + Cr⁺³Cl₃⁻ + K⁺Cl⁻ + H₂⁺O²⁻
1 | 2 | Sn⁺² - 2e = Sn⁺⁴ | ок-е, вос-ль
2 | 4 | Cr⁺⁷ + 4e = Cr⁺³ | вос-е, ок-ль
3P2O3 +4 HNO3 + 7H2O = 6H3PO4 + 4NO
P(+3)-2e = P(+5) 6 в-ль, ок-ие
N(+5)+3e = N(+2) 4 ок-ль, в-ие
дано
m(Mg) = 7.2 g
V(CL2) = 7.8 L
-----------------
m(MgCL2)=?
M(Mg) = 24 g/mol
n(Mg) = m/M = 7.2 / 24 = 0.3 mol
n(CL2) = V(CL2) / Vm = 7.8 / 22.4 = 0.35 mol
n(Mg) < n(CL2)
Mg+CL2-->MgCL2
n(Mg) = n(MgCL2) = 0.3 mol
M(MgCL2) = 95 g/mol
m(MgCL2) = n*M =0.3*95 = 28.5 g
ответ 28.5 г
2)
дано
m(ppa Na2CO3) = 120 g
W(Na2CO3) = 40 %
m(HCL) = 35 g
------------------------
V(CO2)-?
m(Na2CO3) = 120 * 40% / 100% = 48 g
M(Na2CO3) = 106 g/mol
n(Na2CO3) = m/M = 48 / 106 = 0.45 mol
M(HCL) = 36.5 g/mol
n(HCL) = m/M = 35 / 36.5 = 0.99 mol
n(Na2CO3) < n(HCL)
Na2CO3+2HCL-->2NaCL+H2O+CO2
n(Na2CO3) = n(CO2) = 0.45 mol
V(CO2) = n*Vm = 0.45 * 22.4 = 10.08 L
ответ 10.08 л
Дано : m(CuS)=19,2г
Найти:V(H2S)
Решение:
H2S+CuSO4=CuS +H2SO4
n(CuSO4)=19,2/160=0,12моль
n(H2S)=n(CuSO4)=0.12моль
V(H2S)22,4*0,12=2,688л
Ответ:2,688л