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vlpgld7
@vlpgld7
November 2021
1
8
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Помогите срочно 3,4
В 4 там 8%прилили избыток раствора
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Alexei78
3)
CaCO3-->CaO+CO2
CaO+2HNO3-->Ca(NO3)2+H2O
Ca(NO3)2+H2SO4-->CaSO4+2HNO3
Ca⁺²+2NO⁻²₃+2H⁺+SO⁻²₄-->CaSO4+2H⁺+2NO⁻²₃
Ca⁺²+SO⁻²₄-->CaSO4
4)
дано
m(ppa Al2(SO4)3) = 68.4 g
W(Al2(SO4)2) = 8%
-------------------------------
m(BaSO4)-?
m(Al2(SO4)3 = 68.4 * 8% / 100% = 5.472 g
Al2(SO4)3 + 3BaCL2-->3BaSO4+2AlCL3
M(Al2(SO4)3) = 342 g/mol
n(Al2(SO4) 3) = m/M = 5.472 / 342 = 0.016 mol
n(Al2(SO4)3) = 3n(BaSO4)
n(BaSO4) = 3*0.016 = 0.048 mol
M(BaSO4) = 233 g/mol
m(BaSO4) = n*M = 0.048 * 233 = 11.184 g
ответ 11.184 г
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Answers & Comments
CaCO3-->CaO+CO2
CaO+2HNO3-->Ca(NO3)2+H2O
Ca(NO3)2+H2SO4-->CaSO4+2HNO3
Ca⁺²+2NO⁻²₃+2H⁺+SO⁻²₄-->CaSO4+2H⁺+2NO⁻²₃
Ca⁺²+SO⁻²₄-->CaSO4
4)
дано
m(ppa Al2(SO4)3) = 68.4 g
W(Al2(SO4)2) = 8%
-------------------------------
m(BaSO4)-?
m(Al2(SO4)3 = 68.4 * 8% / 100% = 5.472 g
Al2(SO4)3 + 3BaCL2-->3BaSO4+2AlCL3
M(Al2(SO4)3) = 342 g/mol
n(Al2(SO4) 3) = m/M = 5.472 / 342 = 0.016 mol
n(Al2(SO4)3) = 3n(BaSO4)
n(BaSO4) = 3*0.016 = 0.048 mol
M(BaSO4) = 233 g/mol
m(BaSO4) = n*M = 0.048 * 233 = 11.184 g
ответ 11.184 г