дано
m(NaOH) = 50 g
+H2SO4
η(Na2SO4) = 90%
m пр (Na2SO4)-?
2NaOH+H2SO4-->Na2SO4+2HOH
M(NaOH) = 40 g/mol
n(NaOH) = m/M = 50 / 40 = 1.25 mol
2n(NaOH) = n(Na2SO4)
n(Na2SO4) = 1.25 / 2 = 0.625 mol
M(Na2SO4) = 142 g/mol
m теор (Na2SO4) = n*M = 0.625 * 142 = 88.75 g
m пр (Na2SO4) = 88.75 * 90% / 100% = 79.875 g
ответ 79.875 г
Объяснение:
дано
m(NaOH) = 50 g
+H2SO4
η(Na2SO4) = 90%
m пр (Na2SO4)-?
2NaOH+H2SO4-->Na2SO4+2HOH
M(NaOH) = 40 g/mol
n(NaOH) = m/M = 50 / 40 = 1.25 mol
2n(NaOH) = n(Na2SO4)
n(Na2SO4) = 1.25 / 2 = 0.625 mol
M(Na2SO4) = 142 g/mol
m теор (Na2SO4) = n*M = 0.625 * 142 = 88.75 g
m пр (Na2SO4) = 88.75 * 90% / 100% = 79.875 g
ответ 79.875 г
Объяснение: