calculate the volume and number of molecules of CO2 liberated at STP.

1.	calculate the volume and number of molecules of CO2 liberated at STP. If caco3 treated with 36.5g of HCL .( Ca40u,0=16u,c=12u,cl= 35.5)
Answer» 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 :- Let's balance the given CHEMICAL reaction at first! ${\pink{\underline{\boxed{\sf{ CaCO₃ + 2HCl ➞ CaCl₂ + H₂O + CO₂}}}\bigstar}} \; \; \; \; \;\;\;\;$ Here, It is clear that:- 2 mole of HCL when reacted with 1 mole of CaCO₃ gives 1 mole of CO₂.Molar Mass of HCL = 35.5 + 1 = 36.5 g ⠀⠀ $\longrightarrow \sf Num_{(mole)} = Mass given / Molar mass\\$ ⠀⠀ $\longrightarrow \sf Num_{( mole)} = \dfrac{36.5 }{36.5 }\\$ ⠀⠀ $\purple{\longrightarrow \sf Num_{( mole)} = 1 mol}\\$ 2 mole of HCl produces 1 mole of CO₂ so 1 mole of HCl will produce 0.5 mole of CO₂ ⠀⠀ $\longrightarrow \sf Num_{(moles)} = \dfrac{ Volume}{ 22.4}\\$ ⠀⠀ $\longrightarrow \sf 0.5 = \dfrac{ Volume }{22.4}\\$ ⠀⠀ $\longrightarrow \sf Volume = 22.4 \times 0.5\\$ ⠀⠀ $\purple{\longrightarrow \sf Volume = 11.2 litres}\\$ ⠀⠀ $\longrightarrow \sf Num_{( molecules)} = moles × Avogadro Number\\$ ⠀⠀ $\longrightarrow \sf Num_{( molecules)} = 0.5 × 6.022 × 10²³\\$ ⠀⠀ $\purple{\longrightarrow \sf Num_{(molecules)} = 3.011 × 10²³ molecules}\\$ Hence, Volume = 11.2 litres Number of molecules = 3.011 × 10²³ ⠀⠀⠀⠀¯‎¯‎‎‎¯‎¯¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎¯‎‎‎¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎‎¯‎‎¯‎‎‎¯‎‎¯‎¯‎‎‎¯‎‎‎¯‎¯‎¯‎‎¯‎¯‎‎¯‎¯¯‎¯‎¯‎‎‎¯‎¯‎‎‎¯‎‎‎¯‎‎‎¯‎‎‎¯‎