What volume of carbon dioxide is produced when 56.5 g of calcium carbonate react

Question

What volume of carbon dioxide is produced when 56.5 g of calcium carbonate reacts completely according to the following reaction at 25 oC and I atm? calcium carbonate (s)acum oxide (s)+ carbon dioxide(g) iters carbon dioxide Sodium metal reacts with water to produce hydrogen gas according to the following equation: 2Na+2H002NaOH(aq)+Hyg) The product gas,H, is collected over water at a temperature of 25 °C and a pressure of 747 mm Hg. If the wet H, gas formed occupies a volume of 5.77 L, the number of moles of Na reacted was mol. The vapor pressure of water is 23.8 mm Hg at 25 °C. Submit Answer Use the References to access important values if needed for this question. A mixture of helium and oxygengases, in a 5.51 L flask at 83 °C, contains 0.538 grams of helium and 2.94 grams of oxygen. The partial pressure of oxygen in the flask is atm and the total pressure in the flaskis atm. Submit Answer

Explanation / Answer

Write the balanced chemical equation for the reaction.

CaCO3 (s) -------> CaO (s) + CO2 (g)

As per the stoichiometry of the reaction,

1 mole CaCO3 = 1 mole CO2.

Molar mass of CaCO3 = (1*40.078 + 1*12.011 + 3*15.999) g/mol = 100.086 g/mol.

Mole(s) of CaCO3 corresponding to 56.5 g CaCO3 = mole(s) of CO2 produced = (56.5 g)/(100.086 g/mol) = 0.5645 mole.

Assume the CO2 gas to behave ideally and apply the ideal gas law as

P*V = n*R*T where P = 1 atm and T = 25°C = (25 + 273) K = 298 K.

Plug in values and obtain

(1 atm)*V = (0.5645 mole)*(0.082 L-atm/mol.K)*(298 K)

=====> V = (0.5645 mole)*(0.082 L-atm/mol.K)*(298 K)/(1 atm) = 13.794122 L 13.79 L (ans).

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