Question

Un amestec contine 1.05g de SiO2, 0.69 g de celuloză, 1.82g de carbonat de calciu. Cat la suta este carbonatul de calciu? Rotunjiti numarul obtinut. Multumesc

Answer 1

m amestec = 1,05+0,69+1,82 = 3,56g

m CaCO3=1,82g

%CaCO3=mCaCO3/m amestec*100= 1,82/3,56*100=51,123% (procente masice)

MSiO2=ASi+2AO=28+2*16=60g/mol

n=m/M=1,05/60=0,0175 moli SiO2

(C6H10O5)n=6AC+10AH+5AO=6*12+10*1+5*16=162g/mol

n=m/M=0,69/162=0,004259 moli celuloza

MCaCO3=ACa+AC+3AO=40+12+3*16=100g/mol

n=m/M=1,82/100=0,0182 moli CaCO3

nr de moli amestec = 0,0182+0,004259+0,0175=0,04 moli 

nCaCO3 = 0,0182 moli

%CaCO3=nCaCO3/n amestec *100 = 0,0182/0,04*100 = 1,82/0,04=45,5% (procente molare)

Answer 2

m amestec = m celuloza + m SiO2 + m CaCO3
m amestec = 0,69 + 1.05 + 1,82 = 3,56 g amestec

%CaCO3 = m CaCO3/m amestec * 100 = 1,82/3,56 * 100
=>% CaCO3 = 51,123% ≈ 51%