Using the mass of CaCO3 (0.500 g), what volume of CO2 (in mL) could theoretically be obtained if the reaction progressed to completion. Assume that the pressure in the laboratory was 101.3 kPa and the temperature was 22.5 oC and that the gas was behaving ideally. (R = 8.314 kPa L mol-1 K-1)
Reaction is CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
aCO3 + 2HCl --> CaCl2 + CO2 + H2O?
2020-10-06 5:07 pm
回答 (3)
2020-10-06 6:22 pm
Molar mass of CaCO₃ = (40.1 + 12.0 + 16.0×3) g mol⁻¹ = 100.1 g mol⁻¹
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
Mole ratio CaCO₃ : CO₂ = 1 : 1
Moles of CaCO₃ reacted = (0.500 g) / (100.1 g mol⁻¹) = 0.004995 mol
Moles of CO₂ obtained = 0.004995 mol
Consider the CO₂ gas obtained:
Pressure, P = 101.3 kPa
Volume, V = ? mL
No. of moles, n = 0.004995 mol
Gas constant, R = 8.314 kPa L mol⁻¹ K⁻¹ = 8314 kPa mL mol⁻¹ K⁻¹
Temperature, T = (273.2 + 22.5) K = 295.7 K
Gas law: PV = nRT
Then, V = nRT/P
Volume of CO₂ obtained theoretically, V
= 0.004995 × 8314 × 295.7 / 101.3 mL
= 121 mL
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
Mole ratio CaCO₃ : CO₂ = 1 : 1
Moles of CaCO₃ reacted = (0.500 g) / (100.1 g mol⁻¹) = 0.004995 mol
Moles of CO₂ obtained = 0.004995 mol
Consider the CO₂ gas obtained:
Pressure, P = 101.3 kPa
Volume, V = ? mL
No. of moles, n = 0.004995 mol
Gas constant, R = 8.314 kPa L mol⁻¹ K⁻¹ = 8314 kPa mL mol⁻¹ K⁻¹
Temperature, T = (273.2 + 22.5) K = 295.7 K
Gas law: PV = nRT
Then, V = nRT/P
Volume of CO₂ obtained theoretically, V
= 0.004995 × 8314 × 295.7 / 101.3 mL
= 121 mL
2020-10-06 6:28 pm
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
moles of CaCO3 = 0.500 g / 100 g/ mol = 5.00 * 10 ^-3 moles
according to our equation these produce 5.00 * 10 *-3 moles of CO2
volume of these moles
=nRT / P
= 5.00* 10 ^-3 * 8.314 kPa L mol-1 K-1 * (273+22.5) K / 101.3 kPa
= 0.121 L or 121 mL of CO2 under the conditions
moles of CaCO3 = 0.500 g / 100 g/ mol = 5.00 * 10 ^-3 moles
according to our equation these produce 5.00 * 10 *-3 moles of CO2
volume of these moles
=nRT / P
= 5.00* 10 ^-3 * 8.314 kPa L mol-1 K-1 * (273+22.5) K / 101.3 kPa
= 0.121 L or 121 mL of CO2 under the conditions
2020-10-06 8:22 pm
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
From the equation 1 mole of CaCO₃ will produce 1 mole of CO₂
∴
n( CO₂ ) = n( CaCO₃ )
n( CO₂ ) = 0.500g / ( 100 g/mol ) = 0.00500 mol
PV = nRT
V = nRT / P
V = ( 0.00500 mol )( 8.314 kPa L mol⁻¹ K⁻¹ )( 22.5 + 273.15 )K / ( 101.3kPa )
V = 0.121L
V = 121 mL
From the equation 1 mole of CaCO₃ will produce 1 mole of CO₂
∴
n( CO₂ ) = n( CaCO₃ )
n( CO₂ ) = 0.500g / ( 100 g/mol ) = 0.00500 mol
PV = nRT
V = nRT / P
V = ( 0.00500 mol )( 8.314 kPa L mol⁻¹ K⁻¹ )( 22.5 + 273.15 )K / ( 101.3kPa )
V = 0.121L
V = 121 mL
收錄日期: 2021-04-24 08:01:04
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