Răspuns:
MMgCO3=14+12+3×16=84g/mol
MMgO=24+16=40g/mol
84g................40g
MgCO3 => MgO + CO2
2400g..............xg
x=2400×40/84=calculează
n=m/M=2400/84=28,57 moli MgCO3.
1 mol...............1mol
MgCO3 => MgO + CO2
28,57 moli....y moli
y=28,57×1/1=28,57 moli MgO
