Question

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 1

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

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Answer 2

Given :

• CO₂ is liberated at STP if CaCO3 is treated with 36.5 g of HCI.

To Find :

• Volume and number of molecules of CO₂

Solution :

• Let's balance the given chemical reaction at first!

CaCO3 + 2HCI-Ca Cl₂ + H₂O + CO₂

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

Num (mole) Mass given/Molar mass 36.5

→ Num(mole) -

36.5

Num (mole) = 1mol

• 2 mole of HCI produces 1 mole of CO2 so 1 mole of HCI will produce 0.5 mole of CO₂

Num (moles)=Volume/22.4

→ 0.5 =Volume/22.4

Volume = 22.4 x 0.5

→ Volume = 11.2litres

Num(molecules) = moles x Avogadro Number

→ Num (molecules) 0.5 x 6.022 x 10²³

→ Num (molecules) = 3.011× 10²³ molecules

Hence,

• Volume = 11.2 litres

Number of molecules = 3.011 × 10²³