what mass, in grams of nacl formed from the reaction of 2.75ml of a 6.00M Na2SO4 solution and 8.30ml of a 2.50M solution of HCL
Na2SO4 + 2HCl ==> 2NaCl + H2SO4
mols Na2SO4 = M x L = estimated 16 but you do it more accurately.
mols HCl = M x L = estimated 21.
Now use the coefficients in the balanced equation to convert mols of each into mols NaCl.
For Na2SO4 that is estimated 16*2 = estimated 32g
For HCl that is 21 mols x 1 = estimated 21.
You two values for mols NaCl and i is obvious one of them is wrong. In limiting reagent problems the smaller values if ALWAYS the correct value to choose.
Now convert mols NaCl to g. g mols x molar mass.
Well, first things first, we need to figure out the balanced equation for the reaction between Na2SO4 and HCl. But let me put on my lab coat and clown nose for a second...
🤡 Honk honk! 🤡
Okay, let's proceed. The balanced equation can be written as:
Na2SO4 + 2HCl -> 2NaCl + H2SO4
Now let's calculate the moles of Na2SO4 and HCl using the given volumes and concentrations:
Moles of Na2SO4 = Volume x Concentration = (2.75 ml) x (6.00 M) = 16.5 mmol
Moles of HCl = Volume x Concentration = (8.30 ml) x (2.50 M) = 20.75 mmol
According to the balanced equation, the molar ratio between Na2SO4 and NaCl is 1:2. So, the moles of NaCl formed will be twice the moles of Na2SO4:
Moles of NaCl = 2 x Moles of Na2SO4 = 2 x 16.5 mmol = 33 mmol
Now, let's convert moles to grams using the molar mass of NaCl:
Molar mass of NaCl = (22.99 g/mol) + (35.45 g/mol) = 58.44 g/mol
Mass of NaCl = Moles x Molar mass = 33 mmol x 58.44 g/mol = 1926.12 g
So, the mass of NaCl formed from the reaction is approximately 1926.12 grams.
And there you have it, a whole lot of NaCl formed! That's enough salt to season several lifetimes of popcorn. 🍿🧂
To find the mass of NaCl formed from the reaction, you need to determine the limiting reagent first. The limiting reagent is the reactant that is completely consumed and limits the amount of product that can be formed.
Let's start by calculating the moles of Na2SO4 and HCl:
Step 1: Convert the given volumes to liters.
2.75 mL of Na2SO4 solution = 2.75 mL * (1 L / 1000 mL) = 0.00275 L
8.30 mL of HCl solution = 8.30 mL * (1 L / 1000 mL) = 0.00830 L
Step 2: Calculate the moles of Na2SO4 and HCl using the formula: moles = concentration (M) * volume (L).
Moles of Na2SO4 = 6.00 M * 0.00275 L = 0.0165 mol
Moles of HCl = 2.50 M * 0.00830 L = 0.0207 mol
Step 3: Determine the stoichiometry of the balanced chemical equation. The balanced chemical equation for the reaction between Na2SO4 and HCl is:
Na2SO4 + 2HCl -> 2NaCl + H2SO4
According to the equation, one mole of Na2SO4 reacts with 2 moles of HCl to produce 2 moles of NaCl.
Step 4: Determine the limiting reagent. The reactant that has fewer moles is the limiting reagent. In this case, Na2SO4 has 0.0165 moles, and HCl has 0.0207 moles. Since Na2SO4 has fewer moles, it is the limiting reagent.
Step 5: Calculate the moles of NaCl formed. Since Na2SO4 is the limiting reagent, the moles of NaCl formed will be double the moles of Na2SO4 used.
Moles of NaCl formed = 2 * (moles of Na2SO4) = 2 * 0.0165 mol = 0.0330 mol
Step 6: Calculate the mass of NaCl formed using the formula: mass = moles * molar mass.
The molar mass of NaCl is 58.44 g/mol.
Mass of NaCl formed = 0.0330 mol * 58.44 g/mol = 1.9272 g
Therefore, the mass of NaCl formed from the given reaction is approximately 1.93 grams.