Calculate the weight of quicklime formed when 500 g of limestone is heated
(atomic mass of Ca = 40, C = 12, 0 = 16)
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Calculate the weight of quicklime formed when 500 g of limestone is heated
(atomic mass of Ca = 40, C = 12, 0 = 16)
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Verified answer
[tex] \huge \bf \fbox\pink{calc \blue{ulate}}[/tex]
atomic mass of limestone
[tex] CaCO_{3}[/tex]= 40 + 12 + 3×16 = 100 amu
therefore in 1 mole of
[tex] CaCO_{3}[/tex]present in 100 g
therefore 5 moles in 500 g
reaction:
[tex] CaCO_{3}[/tex] (∆)→ CaO +[tex]CO_{2}[/tex]
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
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