Ответ:
дано
m(ppa AgNO3) = 95 g
W(AgNO3) = 4%
m(ppa HCL) = 10 g
W(HCL)=14%
------------------------------
m(AgCL)-?
m(AgNO3)= 95*4% / 100% = 3.8 g
m(HCL) = 10*14%/100% = 1.4 g
M(AgNO3) = 170 g/mol
M(HCL) = 36.5 g/mol
n(AgNO3)= m/M = 3.8/ 170 = 0.022 mol
n(HCL) = m/M= 1.4 /36.5 = 0.038 mol
n(AgNO3)< n(HCL)
3.8 X
AgNO3+HCL-->AgCL↓+HNO3
170 143.5
X = 3.8*143.5 / 170 = 3.2 g
ответ 3.2 г
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Ответ:
дано
m(ppa AgNO3) = 95 g
W(AgNO3) = 4%
m(ppa HCL) = 10 g
W(HCL)=14%
------------------------------
m(AgCL)-?
m(AgNO3)= 95*4% / 100% = 3.8 g
m(HCL) = 10*14%/100% = 1.4 g
M(AgNO3) = 170 g/mol
M(HCL) = 36.5 g/mol
n(AgNO3)= m/M = 3.8/ 170 = 0.022 mol
n(HCL) = m/M= 1.4 /36.5 = 0.038 mol
n(AgNO3)< n(HCL)
3.8 X
AgNO3+HCL-->AgCL↓+HNO3
170 143.5
X = 3.8*143.5 / 170 = 3.2 g
ответ 3.2 г