Question

Se descompun termic 15.5 g CaCO3 de puritate 80% dupa care se adauga o solutie de acid sulfuric de concentratie 49%:
a)masa solutiei de acid
b) numarul de moli de sare rezultat

Answer 1

m CaCO₃ pur = 0,8*15,5 = 12,4g

MCaCO₃ = 100g/mol     ,    n = m/M = 12,4/100 = 0,124 moli CaCO₃

CaCO₃ + H₂SO₄ ⇒ CaSO₄ + CO₂↑+H₂O

a) n CaCO₃ = n H₂SO₄ = 0,124 moli

MH₂SO₄ = 98g/mol ,    n = md/M ⇒ md = n*M = 0,124*98 = 12,152g H₂SO₄

C = md/ms*100 ⇒ ms = md*100/C = 1215,2/49 = 24,8g sol. H₂SO₄

b) nCaCO₃ = n CaSO₄ = 0,124 moli