Reaction: Na2CO3 + 2 HCl -> 2 NaCl + CO2 + H2o
How many grams of NaCl can be produced from the reaction of 1.35g of Na2CO3 with 125mL of 0.265 M HCl?
Another limiting reagent problem. How do I know that? Because amounts are given for BOTH reactrants.
mols Na2COP3 = grams/molar mass = ?
mols HCl = M x L = ?
Using the coefficients in the balanced equation, convert mols Na2CO3 t mols Nacl.
Do the same for mols HCl to mols NaCl.
It is likely that the two values will not be the same which means one of them is wrong; the correct value in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent.
Now convert mols NaCl to g. grams = mols x molar mass. This is the theoretical yield.
To determine the number of grams of NaCl produced from the reaction, we need to follow these steps:
Step 1: Calculate the number of moles of Na2CO3.
Given mass of Na2CO3 = 1.35g
Molar mass of Na2CO3 = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol
Number of moles of Na2CO3 = mass / molar mass
Number of moles of Na2CO3 = 1.35 g / 105.99 g/mol ≈ 0.0127 mol
Step 2: Calculate the number of moles of HCl.
Given volume of HCl = 125 mL = 0.125 L
Molarity of HCl = 0.265 M
Number of moles of HCl = volume (in L) x molarity
Number of moles of HCl = 0.125 L x 0.265 mol/L ≈ 0.0331 mol
Step 3: Use the stoichiometry of the balanced equation to find the number of moles of NaCl.
From the balanced equation: Na2CO3 + 2 HCl -> 2 NaCl + CO2 + H2O
The stoichiometric ratio between Na2CO3 and NaCl is 1:2.
Number of moles of NaCl = 2 x Number of moles of Na2CO3
Number of moles of NaCl = 2 x 0.0127 mol = 0.0254 mol
Step 4: Calculate the mass of NaCl.
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Mass of NaCl = number of moles x molar mass
Mass of NaCl = 0.0254 mol x 58.44 g/mol ≈ 1.484 g
Therefore, approximately 1.484 grams of NaCl can be produced from the reaction of 1.35 grams of Na2CO3 with 125 mL of 0.265 M HCl.
To determine the grams of NaCl produced from the reaction, we need to first calculate the number of moles of Na2CO3 and HCl, and then use the balanced chemical equation to determine the mole ratio between Na2CO3 and NaCl.
1. Calculate the number of moles of Na2CO3:
- Given mass of Na2CO3 = 1.35 g
- Molecular weight of Na2CO3 = 22.99 g/mol (Na) + 12.01 g/mol (C) + 3 * 16.00 g/mol (O) = 105.99 g/mol
- Number of moles of Na2CO3 = given mass / molecular weight = 1.35 g / 105.99 g/mol
2. Calculate the number of moles of HCl:
- Given volume of 0.265 M HCl = 125 mL
- Convert the volume from mL to L: 125 mL x (1 L / 1000 mL)
- Molarity of HCl = 0.265 mol/L
- Number of moles of HCl = molarity x volume = 0.265 mol/L x calculated volume in liters
3. Determine the mole ratio between Na2CO3 and NaCl from the balanced chemical equation:
- From the balanced equation: 1 mole of Na2CO3 produces 2 moles of NaCl
4. Calculate the number of moles of NaCl produced:
- Number of moles of NaCl = number of moles of Na2CO3 x (2 moles of NaCl / 1 mole of Na2CO3)
5. Convert the moles of NaCl to grams:
- Molecular weight of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol
- Mass of NaCl produced = number of moles of NaCl x molecular weight of NaCl
Following these steps, you can calculate the grams of NaCl produced from the given reaction using the given quantities of Na2CO3 and HCl.