help me find which reactant is the limiting reactant when 74 g NaOH and 44 g CO2 are allowed to react when the equation is (2NaOH+1CO2=Na2CO3+1H2O)
1. Convert 74 g NaOH to mols. mols = grams/molar mass
2. Convert 44 g CO2 to mols. Same rocess.
3a. Using the coefficients in the balanced equation convert mols NaOH to mols of either product.
3b. Do the same for mols CO2 to the same produce.
3c. You will have two numbers for mols of the product and both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
This is the long way of doing it but I like it better than the short way.
To determine the limiting reactant, you need to compare the number of moles of each reactant and their stoichiometric ratio in the balanced chemical equation. Here's how you can do it step by step:
1. Calculate the molar mass of NaOH and CO2:
- The molar mass of NaOH (sodium hydroxide) = atomic mass of Na (22.99 g/mol) + atomic mass of O (16.00 g/mol) + atomic mass of H (1.01 g/mol)
= 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
- The molar mass of CO2 (carbon dioxide) = atomic mass of C (12.01 g/mol) + 2 * atomic mass of O (16.00 g/mol)
= 12.01 g/mol + 2 * 16.00 g/mol = 44.01 g/mol
2. Determine the number of moles for each reactant:
- Moles of NaOH = mass of NaOH / molar mass of NaOH
= 74 g / 40.00 g/mol
= 1.85 mol (rounded to two decimal places)
- Moles of CO2 = mass of CO2 / molar mass of CO2
= 44 g / 44.01 g/mol
= 1.00 mol
3. Find the stoichiometric ratio between NaOH and CO2 based on the balanced equation:
The balanced equation is 2NaOH + 1CO2 → Na2CO3 + 1H2O
According to the stoichiometry, 2 moles of NaOH react with 1 mole of CO2 to form 1 mole of Na2CO3.
4. Compare the moles of NaOH and CO2 with their stoichiometric ratio:
- For NaOH, we have 1.85 moles.
- For CO2, we have 1.00 mole.
Since the stoichiometric ratio between NaOH and CO2 is 2:1, we can see that there are more moles of NaOH than CO2. Therefore, CO2 is the limiting reactant because it will be completely consumed in the reaction, while NaOH will be left over.