дано
m(ppa AgNO3) = 95 g
W(AgNO3) = 4%
m(ppa HCL) = 10 g
W(HCL)=14%
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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
3. 2 г