What volume of 0.109 M nitric acid is required to neutralise 2.50g barium hydroxide?
Since the reaction is
2HNO3 + Ba(OH)2 -> Ba(NO3)2 + 2H2O
It takes 2 moles of HNO3 to neutralize 1 mole of Ba(OH2)
2.5g Ba(OH)2 is 0.0146 moles
So, you need 0.0292 moles of HNO3
0.0292M / 0.109M/L = 0.268L of acid.
Well, that sounds like some chemistry fun! Let's calculate it and put a smile on our faces.
First, we need to determine the number of moles of barium hydroxide (Ba(OH)2) in 2.50 grams. To do this, we divide the mass by the molar mass of Ba(OH)2.
The molar mass of Ba(OH)2 is calculated as follows:
1 Ba atom: 137.33 g/mol
2 O atoms: 16.00 g/mol each
2 H atoms: 1.01 g/mol each
Therefore, the molar mass of Ba(OH)2 is:
137.33 + (2 * 16.00) + (2 * 1.01) = 171.35 g/mol
So, the number of moles of Ba(OH)2 is:
2.50 g / 171.35 g/mol = 0.0146 mol
Now, we need to determine the volume of 0.109 M nitric acid (HNO3) required to neutralize this amount of Ba(OH)2. According to the balanced equation:
2 HNO3 + Ba(OH)2 -> 2 H2O + Ba(NO3)2
We can see that the stoichiometric ratio between HNO3 and Ba(OH)2 is 2:1. So, we need half the number of moles of HNO3 as compared to Ba(OH)2.
Therefore, the number of moles of HNO3 required is:
0.0146 mol / 2 = 0.0073 mol
To find the volume, we divide the number of moles by the molarity (concentration) of HNO3:
Volume = moles / Molarity
Volume = 0.0073 mol / 0.109 mol/L = 0.067 L
So, the volume of 0.109 M nitric acid needed to neutralize 2.50g of barium hydroxide (Ba(OH)2) is approximately 0.067 liters.
I hope this chemistry clowning around made the calculations a little more amusing!
To determine the volume of 0.109 M nitric acid required to neutralize 2.50 g of barium hydroxide, we can follow these steps:
Step 1: Determine the molar mass of Barium Hydroxide (Ba(OH)2):
- The molar mass of Ba: 137.33 g/mol
- The molar mass of O: 16.00 g/mol
- The molar mass of H: 1.01 g/mol
- Multiply the molar mass of oxygen by 2 and hydrogen by 2, as there are two of each in the formula.
- Molar mass of Ba(OH)2 = (137.33 g/mol) + (2 * 16.00 g/mol) + (2 * 1.01 g/mol)
- Molar mass of Ba(OH)2 = 171.34 g/mol
Step 2: Calculate the number of moles of barium hydroxide:
- Use the formula: moles = mass / molar mass
- Moles = 2.50 g / 171.34 g/mol
- Moles ≈ 0.0146 mol
Step 3: Based on the balanced equation between nitric acid (HNO3) and barium hydroxide (Ba(OH)2), we can see that they react in a 2:2 ratio:
HNO3 + Ba(OH)2 → Ba(NO3)2 + H2O
This means that 2 moles of HNO3 are required for every 2 moles of Ba(OH)2.
Step 4: Determine the volume of nitric acid required:
- Since the concentration of nitric acid is given as 0.109 M, this means there are 0.109 moles of nitric acid present in one liter of the solution.
- Use the formula: Volume (L) = moles / Molarity
- Volume (L) = (0.0146 mol barium hydroxide) / (0.109 mol/L nitric acid)
- Volume (L) ≈ 0.134 L or 134 mL
Therefore, approximately 134 mL of 0.109 M nitric acid is required to neutralize 2.50 g of barium hydroxide.
To find the volume of the nitric acid required to neutralize the barium hydroxide, we need to use the equation:
Molarity of acid × Volume of acid = Molarity of base × Volume of base
Let's break down the given information:
Molarity of the acid = 0.109 M (given)
Mass of the base = 2.50 g (given)
Molar mass of barium hydroxide (Ba(OH)2) = 137.33 g/mol
First, calculate the number of moles of the base:
Number of moles = Mass / Molar mass
Number of moles of barium hydroxide = 2.50 g / 137.33 g/mol
Next, determine the number of moles of the acid based on the balanced chemical equation. The balanced equation for the reaction of nitric acid (HNO3) and barium hydroxide (Ba(OH)2) is:
2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O
From the balanced equation, we can see that the stoichiometric coefficient of nitric acid is 2. This means that the number of moles of nitric acid required is twice the number of moles of barium hydroxide:
Number of moles of nitric acid (HNO3) = 2 × Number of moles of barium hydroxide
Now that we have the number of moles of acid and base, we can rearrange the equation mentioned earlier to find the volume of the acid:
Volume of acid = (Molarity of base × Volume of base) / Molarity of acid
Volume of acid = (Molarity of base × Number of moles of base) / Molarity of acid
Volume of acid = (0.109 M × (2.50 g / 137.33 g/mol)) / 2
By substituting the given values into the equation and performing the calculations, you can find the volume of nitric acid required to neutralize the barium hydroxide.