Question

When a given mass of Caco3 was heated 0.25dm3 of a gas was collected at 250 and at a pressure of 120Nm-2
1) Write an equation for the reaction
2) Calulate the mass of the gaseous product obtained from (ii) Caco3

Answer 1

Answer on Question #85439 – Chemistry – Other

Task:

When a given mass of CaCO₃ was heated 0.25 dm³ of a gas was collected at 250 and at a pressure of 120 N m⁻²

1) Write an equation for the reaction.

2) Calculate the mass of the gaseous product obtained from (ii) CaCO₃.

Solution:

At 250??? (suppose it is 250 degrees Celsius, 250°C).

Then,

T = 250°C + 273.15 = 523.15 K

P = 120 N m⁻² = 120 Pa

V = 0.25 dm³ = 0.00025 m³

M(CO₂) = Ar(C) + 2*Ar(O) = 12 + 2*16 = 44 (g/mol).

1) Chemical reaction equation:

$$\mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}$$

2) Ideal gas law (Mendeleev-Clapeyron equation):

$$PV = nRT;$$ $$PV = \frac{mRT}{M}$$

where

P – gas pressure (in Pascal);

V – gas volume (in m³);

T – gas temperature (in Kelvins);

R – gas constant 8.314 (m³·Pa·K⁻¹mol⁻¹)

$$m(CO_2) = \frac{PVM}{RT};$$ $$m(CO_2) = \frac{120\,Pa \cdot 0.00025\,m^3 \cdot 44\,g/mol}{8.314\,m^3 \cdot Pa \cdot K^{-1}\text{mol}^{-1} \cdot 523.15\,K} = 0.0003\,g = 0.3\,mg$$ $$m(CO_2) = 0.3\,mg$$

Answer: 0.3 mg CO₂.

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