Sodium metal reacts violently with water to form NaOH and release hydrogen gas. Suppose that 3.00 g of Na reacts completely with 3.00 L of water and the final volume of the system is 3 L. What is the molarity M of the NaOH Solution formed by the reaction?
2Na + 2H2O ==> H2 + 2NaOH
mols Na = 3/23 = estimated 0.13
mols NaOH formed = 0.13mols Na x (2 mols NaOH/2 mol Na) = 0.13 x 2/2 = 0.13 mols NaOH.
M NaOH = mols NaOH/L H2O
Substitute and solve for M.
Oh, chemistry jokes! They never get old! Now let's dig into this reaction equation. We started with 3.00 grams of Na, and if it reacts completely, it's going to produce NaOH.
To find the molarity, we need to calculate the number of moles of NaOH formed, and then divide it by the final volume of the solution.
Step 1: Calculate the number of moles of NaOH formed
The molar mass of Na is 22.99 g/mol. So, 3.00 g of Na would be (3.00 g)/(22.99 g/mol) = 0.1305 mol of Na.
Since the reaction is 1:1 between Na and NaOH, we will have the same number of moles of NaOH as we have of Na. Therefore, we have 0.1305 mol of NaOH formed.
Step 2: Calculate the molarity (M)
The molarity (M) is defined as moles of solute divided by the volume in liters. We have 0.1305 mol of NaOH formed, and the final volume is 3.00 L.
M = (0.1305 mol)/(3.00 L) ≈ 0.0435 M
So, the molarity of the NaOH solution formed by the reaction is approximately 0.0435 M. And remember, if you ever feel stressed in chemistry class, just remember to stay positive!
To find the molarity (M) of the NaOH solution formed by the reaction, we first need to calculate the number of moles of Na used and then use it to determine the molarity of the NaOH.
1. Calculate the number of moles of Na:
Given the mass of Na is 3.00 g, we can use the molar mass of Na to find the number of moles. The molar mass of Na is 22.99 g/mol.
Moles of Na = Mass of Na / Molar mass of Na = 3.00 g / 22.99 g/mol ≈ 0.1304 moles of Na
2. Calculate the molarity of the NaOH solution:
Since Na reacts with water to form NaOH, we can assume that the moles of Na will be equal to the moles of NaOH formed.
Molarity (M) = Moles of solute (NaOH) / Volume of solution (in liters)
Molarity = Moles of NaOH / Volume of solution
The volume of the solution is the final volume of the system, which is given as 3 L.
Therefore, Molarity (M) = 0.1304 moles of NaOH / 3 L ≈ 0.0435 M
So, the molarity of the NaOH solution formed by the reaction is approximately 0.0435 M.
To find the molarity (M) of the NaOH solution formed by the reaction, we need to determine the number of moles of NaOH in the final solution.
First, we need to find the number of moles of Na reacted. Since the molar mass of Na is 22.99 g/mol, we can calculate the moles as follows:
moles of Na = mass of Na / molar mass of Na
moles of Na = 3.00 g / 22.99 g/mol
moles of Na ≈ 0.1305 mol
Next, we need to determine the moles of NaOH formed. From the balanced chemical equation, we know that 2 moles of Na reacts to produce 2 moles of NaOH:
2Na + 2H2O → 2NaOH + H2
Since the reaction is stoichiometrically balanced, the moles of NaOH formed will be the same as the moles of Na reacted:
moles of NaOH = moles of Na ≈ 0.1305 mol
Now, we can calculate the molarity of the NaOH solution by dividing the moles of NaOH by the final volume of the system in liters:
Molarity (M) = moles of NaOH / volume of solution (in L)
Molarity (M) = 0.1305 mol / 3 L
Molarity (M) ≈ 0.0435 M
Therefore, the molarity of the NaOH solution formed by the reaction is approximately 0.0435 M.