Answer:
Therefore, volume of CO₂ produced in the first step is 9141.404 L
Explanation:
Equations of reactions:
A: CaCO₃(s) ---> CaO(s) + CO₂(g)
B: CaO(l) + H₂O(l) ---> Ca(OH)₂(s)
Molar mass of CaCO₃ = 100 g; molar mass of CaO = 56 g; molar mass of CO₂ = 44 g molar mass of H₂P = 18 g; molar mass of Ca(OH)₂ = 74 g
From equation B, 1 mole of CaO produces 1 mole of Ca(OH)₂
This means that 56 g of CaO produces 74 g of Ca(OH)₂
mass of CaO that produces 8.47 kg or 8470 g of Ca(OH)₂ = 8470 g * 56/74 = 6409.73 g of CaO
Therefore, 6409.73 g of CaO were produced in reaction A
From reaction A, 1 mole of CaCO₃ produces 1 mole CaO and 1 mole of CO₂
Number of moles of CaO in 6409.73 g = 6409.73 g/56 g/mol = 114.46 moles
Therefore, 114.46 moles of CO₂ were produces as well.
Molar volume of gas at STP = 22.4 litres
Volume of CO₂ produced at STP = 114.46 * 22.4 L =2563.904 L
However, the above reaction took place at 950 K and 0.976 atm, therefore volume of CO₂ produced under these conditions are obtained using the general gas equation
Using P₁V₁/T₁ = P₂V₂/T₂
P₁ = 1.0 atm, V₁ = 2563.904 L, T₁ = 273 K, P₂ = 0.976 atm, T₂ = 950 K, V₂ = ?
V₂ = P₁V₁T₂/P₂T₁
V₂ = (1.0 * 2563.904 * 950)/(0.976 * 273)
V₂ = 9141.404 L
Therefore, volume of CO₂ produced in the first step is 9141.404 L
