Răspuns:
CuCl₂+ 2 NaOH →Cu(OH)₂ + 2NaCl
AlCl₃ + 3 NaOH → Al(OH)₃ + 3 NaCl
Cu(OH)₂ hidroxid de cupru
M,Cu(OH)₂ =64+2ˣ(16+1)= 98g/mol
M, Al(OH)₃ = 27+3ˣ(16+1) = 78g/mol
Raport atomic
Cu÷O÷ H=1÷2÷2
Al÷O÷H = 1÷3÷3
Raport masic
mCu÷mO÷mH= 64÷32÷2 simplificăm cu 2
mCu÷mO÷mH= 32÷16÷1
mAl÷mO÷mH =27÷ 48÷ 3 simplicăm cu 3
mAl÷mO÷mH =9÷ 16÷ 1
AlCl₃ + 3 NaOH → Al(OH)₃ + 3 NaCl
M,NaOH= 23+16+1=40g/mol
3 ˣ40 g NaOH.......................78gAl(OH)₃
80 g NaOH.................................X
X=(80 ˣ78)÷(3 ˣ 40)=52g Al(OH)₃
Explicație:
