Question

cati moli se gasesc in: 0,8 kg carbonat de Ca
126 kg acid azotic
1260 g acid sulfuric
Ce cantitate de substanta pura se gaseste in 400 g calcar cu 5% impuritati?

Answer 1

MCaCO3 = 100g/mol

m = 0,8kg = 800g CaCO3

n = m/M = 800/100 = 8 moli CaCO3

MHNO3 = 63g/mol

m = 126 kg = 126000g moli

n = m/M = 126000/63 = 2000 moli HNO3

MH2SO4 = 98g/mol

n = m/M = 1260/98 = 12,857 moli H2SO4

p = 100-5 = 95%

p = m pur/m impur*100 => m pur = p*m impur/100 = 95*400/100 = 380 CaCO3