For the reaction:
2HCl + Na2CO3→ 2NaCl + H2O + CO2
how much CO2 is produced at STP if 18.5 L of 1.15 M Na2CO3is used? The molar mass of Na2CO3is 106.01 g/mol.
I know I have to use stoichiometry and possibly PV=nRT but I don't know what to use first. Do I start with 18.5L?
whta's the answer?
the answer would be 477 liters
To determine the amount of CO2 produced at STP, you can use stoichiometry and the ideal gas law equation (PV = nRT). Here's how you can approach the problem:
Step 1: Write and balance the chemical equation:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
Step 2: Determine the number of moles of Na2CO3:
To do this, use the given volume and molarity of Na2CO3. The molarity (M) of a solution is defined as the number of moles of solute per liter of solution.
Given:
Volume of Na2CO3 solution (V) = 18.5 L
Molarity of Na2CO3 solution (M) = 1.15 M
Using the formula:
moles of solute = volume of solution (in liters) × molarity
Number of moles of Na2CO3 = 18.5 L × 1.15 mol/L
Step 3: Calculate the number of moles of CO2 using stoichiometry:
From the balanced chemical equation, we can see that the stoichiometric ratio between Na2CO3 and CO2 is 1:1. Therefore, the number of moles of CO2 produced will be equal to the number of moles of Na2CO3 used.
Number of moles of CO2 = Number of moles of Na2CO3 (from step 2)
Step 4: Convert moles of CO2 to grams:
To convert moles to grams, you'll need the molar mass of CO2, which is 44.01 g/mol.
Mass of CO2 = Number of moles of CO2 (from step 3) × molar mass of CO2
Step 5: Convert grams of CO2 to volume at STP:
At STP (standard temperature and pressure), 1 mole of any ideal gas occupies 22.4 L of volume.
Volume of CO2 at STP = Mass of CO2 (from step 4) / molar mass of CO2 × 22.4 L/mol
Following these steps should enable you to calculate the amount of CO2 produced in the given reaction.