You weigh out 1.20 g of sodium carbonate monohydrate, dissolve it in water and react it with 15.00 mL of 2.00 M HCl. Write an equation for the reaction and calculate the expected mass of carbon dioxide that will be produced.
This is a limiting reagent problem. I know that because amounts are given for BOTH reactants.
Na2CO3 + 2HCl ==> H2O + CO2 + 2NaCl
mols Na2CO3 = grams/molar mass = ?
mols HCl = m x L = ?
Convert mols Na2CO3 to mols CO2. use the coefficients in the balanced equation to do this.
Convert mols HCl to mols CO2 the same way.
It is likely that the mols from these steps will not agree and one of them must not be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Use the smaller value and convert to grams CO2. g = mols x molar mass.
The equation for the reaction between sodium carbonate monohydrate (Na2CO3·H2O) and hydrochloric acid (HCl) can be written as follows:
Na2CO3·H2O + 2HCl -> 2NaCl + H2O + CO2
According to the balanced equation, one mole of sodium carbonate monohydrate reacts with 2 moles of hydrochloric acid to produce 1 mole of carbon dioxide gas.
To calculate the expected mass of carbon dioxide that will be produced, we need to determine the moles of sodium carbonate monohydrate used and then convert it to moles of carbon dioxide. Finally, we can convert the moles of carbon dioxide to grams using the molar mass of CO2.
First, let's find the moles of sodium carbonate monohydrate:
Given mass = 1.20 g
Molar mass of Na2CO3·H2O = (2*23.00 g/mol) + (12.01 g/mol) + (3*16.00 g/mol) + 2*(1.01 g/mol) + 16.00 g/mol
= 105.99 g/mol
Moles = Mass / Molar mass
= 1.20 g / 105.99 g/mol
= 0.011 mol (approximately)
Since the stoichiometry of the equation shows that 1 mole of Na2CO3·H2O produces 1 mole of CO2, we can conclude that 0.011 moles of Na2CO3·H2O will produce 0.011 moles of CO2.
Finally, let's convert the moles of CO2 to grams:
Molar mass of CO2 = (12.01 g/mol) + 2*(16.00 g/mol)
= 44.01 g/mol
Mass = Moles x Molar mass
= 0.011 mol x 44.01 g/mol
= 0.484 g (approximately)
Therefore, the expected mass of carbon dioxide that will be produced is approximately 0.484 grams.