Question

Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide. Calculate how many grams of quicklime can be produced from 5.0 kg of limestone. × 10 g Enter your answer in scientific notation.

Answer 1

Molar mass of CaCO3 = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

mass of CaCO3 = 5 Kg = 5000 g

mol of CaCO3 = (mass)/(molar mass)

= 5000/100.09

= 49.955 mol

we have the Balanced chemical equation as:

CaCO3 ---> CaO + CO2

From balanced chemical reaction, we see that

when 1 mol of CaCO3 reacts, 1 mol of CaO is formed

mol of CaO formed = moles of CaCO3

= 49.955 mol

Molar mass of CaO = 1*MM(Ca) + 1*MM(O)

= 1*40.08 + 1*16.0

= 56.08 g/mol

mass of CaO = number of mol * molar mass

= 49.955*56.08

= 2801 g

Answer: 2.80*10^3 g