Respuesta :
26.3 grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at 901 mmHg and [tex]0^0C[/tex]
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 901 mm Hg = 1.18 atm (760 mm Hg = 1 atm)
V = Volume of gas = 5.00 L
n = number of moles = ?
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]0^0C=(0+273)K=273K[/tex]
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{1.18atm\times 5.00L}{0.0821L atm/K mol\times 273K}=0.263moles[/tex]
The chemical reaction for the decomposition of calcium carbonate follows the equation:
[tex]CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)[/tex]
According to stoichiometry:
1 mole of carbon dioxide is produced by = 1 mole of calcium carbonate
Thus 0.263 moles of carbon dioxide is produced by = [tex]\frac{1}{1}\times 0.263=0.263[/tex] moles of calcium carbonate
Mass of calcium carbonate=[tex]moles\times {\text {Molar mass}}=0.263mol\times 100g/mol=26.3g[/tex]
26.3 grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at 901 mmHg and [tex]0^0C[/tex]
