Question

Calculate gram moles of $CaO$ that could be obtained from $42.54g$ of \[CaC{O_3}\] and convert the number of g-moles to gram.

Answer 1

Hint: Calcium carbonate undergoes thermal decomposition and forms carbon dioxide and calcium oxide. By constructing a balanced equation of the same a relationship between the moles of Calcium carbonate and Calcium oxide can be established.

Complete step by step answer: \[CaC{O_3}\] , its chemical name is Calcium Carbonate and it is also called as Limestone, Chalk and Marble. Calcium Carbonate undergoes thermal decomposition and gives Calcium Oxide and Carbon dioxide.
$CaC{O_3} \to CaO + C{O_2}$
$\;100g \to 56g + 44g$
$40g\; + 12g + 3 \times 16 = 100g$
Molecular mass o $CaC{O_3} = 100g$
Calcium molecular mass = $40g$
Carbon molecular mass = $12g$
Oxygen = $16g$
Molecular weight of Calcium + Molecular weight of Carbon + $3 \times $Molecular weight of Oxygen
$40g\; + 12g + 3 \times 16 = 100g$
Calcium oxide chemical formula $CaO$ and it is also called as quick lime
Molecular mass of Calcium Oxide = $56g$
Molecular weight of Calcium + Molecular Oxygen
$40g + 16g = 56g$
Mole is defined as heap of particles containing number is equal to $6.023 \times {10^{23}}$
So this number is also called avogadro number.
The formula for finding gram mole is written as $n\; = \dfrac{{weight\;of\;subs\tan ce}}{{Gram\;molecular\;weight}}$
$CaC{O_3} \to CaO + C{O_2}$
$\;100g \to 56g + 44g$
So from these equation we can say

 $100g$ of Calcium carbonate produces $1$ mole of Calcium Oxide
$42.54g$ of Calcium carbonate will gives ----------- X
$X \times 100g\;\operatorname{of} \;calcium\;carbonate\; = 42.5g\;\operatorname{of} \;calcium\;carbonate\; \times 1\;\operatorname{mole} \;of\;Calcium\;oxide$
$X\; = \dfrac{{42.5 \times 1}}{{100}} = 0.425\;gram\;mole$
$1$ gram mole of Calcium oxide is $56g$
$0.425$ gram mole of Calcium oxide = $X$
$X \times \;1\;\operatorname{gram} \;mole\;\,Calcium\;oxide = 56g\;\operatorname{of} \;Calcium\;oxide \times 0.425gram\;mole\;\operatorname{of} \;calcium\;oxide$$X = \dfrac{{56g \times 0.425}}{1} = 23.8g\;\operatorname{of} \;Calcium\;oxide$
So the decomposition of $42.5g$ of Calcium carbonate produces $0.425$ gram moles of Calcium oxide and $0.425$ gram moles of Calcium oxide is equal to $23.8g$ of Calcium oxide.


Note:The molecular weight of Calcium carbonate is $100g$ and molecular weight of Calcium oxide is $56$ grams and $100g$ of Calcium carbonate gives $56g$ of Calcium Oxide.