Question

помогите решить 4 задачу

Answer 1

дано

V(HCL)  = 50 mL

W(HCL) = 5%

p(HCL) = 0.98 g/mL

V(AgNO3) = 40 mL

W(AgNO3) = 10%

p(AgNO3) = 1.02 g/mL

-----------------------------------

m(AgCL)-?

m(HCL) = V(HCL)*W(HCL)*p(HCL) / 100% = 50*5*0.98 / 100 = 2.45g

M(HCL) = 36.5 g/mol

n(HCL) = m/M = 2.45 / 36.5 = 0.07 mol

m(AgNO3) = V(AgNO3) * W(AgNO3) * p(AgNO3) / 100%

m(AgNO3) = 40*10*1.02 / 100 = 4.08 g

M(AgNO3) = 170 g/mol

n(AgNO3) = m/M = 4.08 / 170 = 0.024 mol

n(HCL) > n(AgNO3)

HCL+AgNO3-->AgCL↓+HNO3

n(AgNO3) = n(AgCL) = 0.024 mol

M(AgCL) = 143.5 g/mol

m(AgCL) = n(AgCL) * M(AgCL) = 0.024 * 143.5 = 3.444 g

ответ 3.444 г