A mixture of calcium and magnesium carbonates weighing 1.4 g was strongly heated until no further loss of weight was perceived. The residue weighed 0.76 g. What percentage of MgCO3 was present in the mixture?
A
20.45
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
18.55
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
16.33
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
22.4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A20.45 CaCO3+MgCO3→CaO+MgO
let mass of CaCO3 = x g
and mass of MgCO3 = 1.4−x g
moles of CaCO3 = x100 = moles CaO produced
moles of MgCO3 = x84 = moles MgO produced
mass of CaO produced = moles * no. of moles = (x)∗56100
mass of MgO produced = moles * no. of moles = (1.4−x)∗40100
so, total weight of product = (1.4−x)∗56100 + (1.4−x)∗40100 = 0.76