Question

O masă de 400g de CaCO3 cu puritatea de 40% se descompune termic. Calculeaza masa de oxid metalic obtinut, volumul de gaz rezultat din reactie si puritatea oxidului obtinut.

Answer 1

CaCO3 => CaO + CO2
puritatea = m pura/m impura * 100
m pura = 400 * 40/100 = 160 g CaCO3
100 g CaCO3.......56 g CaO
160 g CaCO3..........x g CaO
x = 89.6 g CaO = m pura CaO

100 g CaCO3...........22.4 l CO2
160 g CaCO3...............x l CO2
x = 35.84 l CO2

puritatea = m pura/m impura * 100
100 g CaCO3............56 g CaO
400 g CaCO3..............x g CaO
x = 224 g CaO = m impura CaO
puritatea = 89.6/224 * 100 = 40%

Answer 2

puritatea = m pur / m impur *100 => mpur=p*m impur/100 = 40*400/100=160g CaCO₃

masa impura => 400-160=240g


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

MCaO = ACa+AO=40+16=56g/mol

100g...................56g
CaCO₃ (t⁰C) =  CaO + CO₂
160g.................x=89,6g (masa pura)

100g CaCO₃.....................22,4L CO₂
160g CaCO₃......................x=35,84L CO₂

100g CaCO₃......................56g CaO
400g CaCO₃.....................x=224g CaO (masa impura)

puritatea = masa pura/ masa impura *100 = 89,6/224*100=40%