Определите, какие массы раствора фосфорной кислоты с массовой частей H3PO4 3%,

дано

W(H3PO4) = 3%

W(KOH) = 5%

m(ppa K3PO4) = 300 g

W(K3PO4) = 2%

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

m(ppa H3PO4)-?

m(ppa KOH)-?

m(K3PO4) = 300 * 2% / 100% = 6 г

H3PO4+ 3KOH--gt;K3PO4+3H2O

M(K3PO4) = 212 g/mol

n(K3PO4) = m/M = 6 / 212 = 0,028 mol

n(H3PO4) = n(K3PO4) = 0.028 mol

M(H3PO4) = 98 g/mol

m(H3PO4) = n*M = 0.028 * 98 = 2.744 g

m(ppa H3PO4) = 2.744 * 100% / 3% =  91.47 g

3n(KOH) = n(K3PO4)

n(KOH) = 3* 0.028  = 0.084 mol

M(KOH) = 56 g/mol

m(KOH) = n*M = 0.084 * 56 = 4.704 g

m(ppa KOH) = 4.704 * 100% / 5% =  94.08 g

ответ 91.47 г ,94 .08 г