Question

Calculate the weight of quicklime formed when 500 g of limestone is heated
(atomic mass of Ca = 40, C = 12, 0 = 16)​

Answer 1

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atomic mass of limestone

$CaCO_{3}$= 40 + 12 + 3×16 = 100 amu

therefore in 1 mole of

$CaCO_{3}$present in 100 g

therefore 5 moles in 500 g

reaction:

$CaCO_{3}$ (∆)→ CaO +$CO_{2}$

atomic mass of CaO = 40 + 16 = 56 amu therefore 5 moles will weigh 280 g

atomic mass of CO2 = 12 + 2×16 = 44 amu

therefore 5 moles will weigh 220 g

therefore mass of CaO formed will be 280 g

Answer 2

Explanation:

\huge \bf \fbox\pink{calc \blue{ulate}}

calculate

atomic mass of limestone

CaCO_{3}CaCO

3

= 40 + 12 + 3×16 = 100 amu

therefore in 1 mole of

CaCO_{3}CaCO

3

present in 100 g

therefore 5 moles in 500 g

reaction:

CaCO_{3}CaCO

3

(∆)→ CaO +CO_{2}CO

2

atomic mass of CaO = 40 + 16 = 56 amu therefore 5 moles will weigh 280 g

atomic mass of CO2 = 12 + 2×16 = 44 amu

therefore 5 moles will weigh 220 g

therefore mass of CaO formed will be 280 g