2H3PO4+3Ca(OH)2=Ca3(PO4)2+6H2O
m(Ca(OH)2)-->v
v(Ca(OH)2)=m(Ca(OH)2)/M
v(Ca(OH)2)=7.4/74=0.1
v(H3PO4)=0.07
m(H3PO4)=v(H3PO4)*M=0.07*98=7
Мr(Сr₂O₃) = 52 * 2 + 16 * 3 = 152
Mr(Cr) = 52 * 2 = 104
w%(Cr) = 104 : 152 * 100 % = 68,42%
2Al + 3F₂ = 2AlF₃
по уравнению реакции
m(Al)/[2M(Al)] = m(AlF₃)/[2M(AlF₃)]
откуда масса алюминия
m(Al)=2M(Al)*m(AlF₃)/[2M(AlF₃)]
m(Al)=M(Al)*m(AlF₃)/[M(AlF₃)]
m(Al)=27г/моль*16,8г/84г/моль=5,4г (ответ)