22.0g of zinc carbonate was heated
vigorously to a constant mass at STP what
volume of carbon dioxide is produced at 800
mmHg and 30°C?(atomic weight of Zn is
65.5g)
Answers
It has given that, 22g of Zinc carbonate was heated vigorously to a constant mass at STP. pressure = 800 mm Hg and temperature = 30°
we have to find volume of carbon dioxide is produced.
solution : see chemical reaction,
ZnCO₃ (zinc carbonate) ⇒ZnO (zinc oxide) + CO₂ (carbon dioxide)
no of moles of Zinc carbonate = no of moles of Carbon dioxide
⇒22g/65.5 g/mol = no of moles of CO₂
⇒no of moles of CO₂ = 0.33587
now using formula, PV = nRT
here P = 800 mm Hg = 800/760 atm = 1.052 atm
T = 30° = 273 + 30 = 303 K
n = 0.33587 and R = 0.082 atm.L/mol/K
so, V = (0.33587 × 0.082 × 303)/(1.052)
= 7.9476 L ≈ 8 Litre
Therefore the volume of carbon dioxide is 8 Litre.
Answer:
Step-by-step explanation:
It has given that, 22g of Zinc carbonate was heated vigorously to a constant mass at STP. pressure = 800 mm Hg and temperature = 30°
we have to find volume of carbon dioxide is produced.
solution : see chemical reaction,
ZnCO₃ (zinc carbonate) ⇒ZnO (zinc oxide) + CO₂ (carbon dioxide)
no of moles of Zinc carbonate = no of moles of Carbon dioxide
⇒22g/65.5 g/mol = no of moles of CO₂
⇒no of moles of CO₂ = 0.33587
now using formula, PV = nRT
here P = 800 mm Hg = 800/760 atm = 1.052 atm
T = 30° = 273 + 30 = 303 K
n = 0.33587 and R = 0.082 atm.L/mol/K
so, V = (0.33587 × 0.082 × 303)/(1.052)
= 7.9476 L ≈ 8 Litre
Therefore the volume of carbon dioxide is 8 Litre.