5g of calcium carbonate reacted with 0.5mols of sulfuric acid, what mass of carbon dioxide is produced?
This is a limiting reagent problem. Here is a worked example. Just follow the steps.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
To find the mass of carbon dioxide produced, we first need to determine the balanced chemical equation for the reaction between calcium carbonate and sulfuric acid:
CaCO3 + H2SO4 -> CaSO4 + CO2 + H2O
From the equation, we can see that 1 mole of calcium carbonate (CaCO3) produces 1 mole of carbon dioxide (CO2). Therefore, we need to calculate the number of moles of calcium carbonate reacted.
Given that 5g of calcium carbonate reacted, we can calculate the number of moles using the molar mass of calcium carbonate. The molar mass of calcium carbonate (CaCO3) is the sum of the atomic masses of its constituent elements:
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol) = 100.09 g/mol
Number of moles = Mass / Molar mass = 5g / 100.09 g/mol = 0.04995 moles (rounded to 5 significant figures)
According to the balanced equation, 1 mole of calcium carbonate produces 1 mole of carbon dioxide. Therefore, the number of moles of carbon dioxide produced is also 0.04995 moles.
To find the mass of carbon dioxide produced, we can use the molar mass of carbon dioxide (CO2), which is 44.01 g/mol:
Mass = Number of moles x Molar mass = 0.04995 moles x 44.01 g/mol = 2.196 g (rounded to 3 decimal places)
Therefore, approximately 2.196 grams of carbon dioxide is produced.