calculate the volume and number of molecules of co2 liberated at STP if caco3 is treated with 36.5 gram of HCL
Answers
Given :-
▪ CO₂ is liberated at STP if CaCO₃ is treated with 36.5 g of HCl
To Find :-
▪ Volume and number of molecules of CO₂
Solution :-
First, Let us write the balanced chemical equation for the given chemical reaction.
CaCO₃ + 2HCl ➞ CaCl₂ + H₂O + CO₂
From the balanced chemical reaction, It is clear that,
2 mole of HCL when reacted with 1 mole of CaCO₃ gives 1 mole of CO₂
We have
- Molar Mass of HCL = 35.5 + 1 = 36.5 g
- Given mass of HCL = 36.5 g
So,
⇒ Num. of mole = Mass given / Molar mass
⇒ Num. of mole = 36.5 / 36.5
⇒ Num. of mole = 1 mol
Since, 2 mole of HCl produces 1 mole of CO₂ so 1 mole of HCl will produce 0.5 mole of CO₂
Now,
⇒ Num. of moles = Volume / 22.4
⇒ 0.5 = Volume / 22.4
⇒ Volume = 22.4 × 0.5
⇒ Volume = 11.2 litres
Also, We know, 1 mole of any molecule has 6.022 × 10²³ molecules so,
⇒ Num. of molecules = moles × Avogadro Number
⇒ Num. of molecules = 0.5 × 6.022 × 10²³
⇒ Num. of molecules = 3.011 × 10²³ molecules
Hence,
- Volume = 11.2 litres
- Number of molecules = 3.011 × 10²³
Explanation:
Given :-
▪ CO₂ is liberated at STP if CaCO₃ is treated with 36.5 g of HCl
To Find :-
▪ Volume and number of molecules of CO₂
Solution :-
First, Let us write the balanced chemical equation for the given chemical reaction.
CaCO₃ + 2HCl ➞ CaCl₂ + H₂O + CO₂
From the balanced chemical reaction, It is clear that,
2 mole of HCL when reacted with 1 mole of CaCO₃ gives 1 mole of CO₂
We have
Molar Mass of HCL = 35.5 + 1 = 36.5 g
Given mass of HCL = 36.5 g
So,
⇒ Num. of mole = Mass given / Molar mass
⇒ Num. of mole = 36.5 / 36.5
⇒ Num. of mole = 1 mol
Since, 2 mole of HCl produces 1 mole of CO₂ so 1 mole of HCl will produce 0.5 mole of CO₂
Now,
⇒ Num. of moles = Volume / 22.4
⇒ 0.5 = Volume / 22.4
⇒ Volume = 22.4 × 0.5
⇒ Volume = 11.2 litres
Also, We know, 1 mole of any molecule has 6.022 × 10²³ molecules so,
⇒ Num. of molecules = moles × Avogadro Number
⇒ Num. of molecules = 0.5 × 6.022 × 10²³
⇒ Num. of molecules = 3.011 × 10²³ molecules
Hence,
Volume = 11.2 litres
Number of molecules = 3.011 × 10²³