1) 35% от 125 грамм = 125*0.35 = 43.75 грамм HCl

2) AgNO3 + HCl = AgCl + HNO3

43.75/36.5 = x/143

x = 171.4 грамма осадка

Ответ:

дано

m(ppa HCL) = 125 g

W(HCL) = 35%

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m(AgCL)-?

m(HCL) = m(ppaHCL) * W(HCL) / 100% = 125 * 35% / 100% = 43.75 g

AgNO3+HCL-->AgCL+HNO3

M(HCL) = 36.5 g/mol

n(HCL) = m(HCL) / M(HCL) = 43.75 / 36.5 = 1.2 mol

n((HCL) = n(AgCL) = 1.2 mol

M(AgCL) = 143.5 mol

m(AgCL) = n(AgCL) * M(AgCL) = 1.2 * 143.5= 172.2 g

ответ 172.2 г

Объяснение: