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Consider the carbon dioxide gas formed in the reaction :
Pressure, P = 734/760 atm
Volume, V = 0.023 m³ = 23 L
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273 + 250) K = 523 K

Gas law : PV = nRT
No. of moles of CO₂ produced = PV/RT = (734/760) × 23 / (0.08206 × 523) mol = 0.518 mol

Ca(HCO₃)₂ would decompose to give CaCO₃, H₂O and CO₂. However, CaCO₃ would further decompose to give CaO and CO₂ in such a high temperature of 250°C. Hence, the equation for the reaction is :
Ca(HCO₃)₂ → CaO + H₂O + 2CO₂
Mole ratio Ca(HCO₃)₂ : CO₂ = 1 : 2
No. of moles of Ca(HCO₃)₂ used = (0.518 mol) × (1/2) = 0.259 mol

Molar mass of Ca(HCO₃)₂ = (40.01 + 1.0×2 + 12.0×2 + 16.0×6) g/mol = 162.1 g/mol
Mass of Ca(HCO₃)₂ used = (162.1 g/mol) × (0.259 mol) = 42.0 g

0.001 meters^3 = 1liter

0.023 meters^3 x 1000 liter/meter^3 = 23 liters

PV = nRT ; R = 0.082 liter-atm/mole K ; T = 250 C + 273 = 523 Kelvin


734 mm x 1 atm/760 mm (23 liters) = n (0.082 liter-atm/mole K(523)

n = 0.517 moles gas

Ca(HCO)3 (s) ==⇒ CaCO3(s) + H2O(g) + CO2(g)

0.517 moles gas x 1mole Ca(HCO3)2 / 2moles gas x 162 grams Ca(HCO3)2/mole Ca(HCO3)2 = 41.9 g Ca(HCO3)2

CaHCO3 -----> CaO + CO2 + 1/2H2