Consider the reaction:
2 NaBH4(aq) + H2SO4(aq) → 2 H2(g) + Na2SO4(aq) + B2H6(g)
What volume, in mL, of a 0.500 M solution of NaBH4 is required to produce 0.529 g of B2H6? H2SO4 is present in excess.
Well, let's break down the problem, shall we?
Given:
- Mass of B2H6 produced: 0.529 g
- Concentration of NaBH4 solution: 0.500 M
First things first, we need to convert the mass of B2H6 to moles. But don't worry, I won't start singing "moles on the water"!
The molar mass of B2H6 is approximately 27.67 g/mol. So, dividing the mass by the molar mass, we find that we have about 0.0191 moles of B2H6.
Now, let's consider the stoichiometry of the reaction. According to the equation, 2 moles of NaBH4 produce 1 mole of B2H6. So, for every 2 moles of NaBH4, we get 0.0191 moles of B2H6.
Since the concentration of the NaBH4 solution is given in moles per liter (M), we can use the equation C = n/V to find the volume.
0.500 M = 0.0191 moles / V (in liters)
Rearranging the equation, we find that V (in liters) = 0.0191 moles / 0.500 M, which gives us approximately 0.0382 liters.
Now, to convert liters to milliliters, we multiply by 1000. So, V (in mL) = 0.0382 L * 1000, which equals around 38.2 mL.
So, you would need approximately 38.2 mL of the 0.500 M NaBH4 solution to produce 0.529 g of B2H6. Now that's a-mole-zing!
To determine the volume of the 0.500 M solution of NaBH4 needed, we can use the molar mass of B2H6 and stoichiometry of the reaction.
1. Calculate the number of moles of B2H6:
- Given mass of B2H6: 0.529 g
- Molar mass of B2H6: 27.7 g/mol
- Moles of B2H6 = mass / molar mass = 0.529 g / 27.7 g/mol = 0.0191 mol
2. Use the stoichiometry of the reaction to find the number of moles of NaBH4:
- From the balanced equation, we see that 1 mole of B2H6 reacts with 2 moles of NaBH4.
- Thus, moles of NaBH4 = 0.0191 mol × (2 mol NaBH4 / 1 mol B2H6) = 0.0382 mol
3. Calculate the volume of the 0.500 M NaBH4 solution required:
- By using the equation: moles = concentration (M) × volume (L)
- Rearrange the equation to solve for volume: volume (L) = moles / concentration (M)
- Volume (L) = 0.0382 mol / 0.500 mol/L = 0.0764 L
4. Convert volume from liters to milliliters:
- Recall that 1 L = 1000 mL
- Volume (mL) = 0.0764 L × 1000 mL/L = 76.4 mL
The volume of the 0.500 M solution of NaBH4 required to produce 0.529 g of B2H6 is 76.4 mL.
To determine the volume of a 0.500 M solution of NaBH4 required to produce 0.529 g of B2H6, we need to use stoichiometry and the molarity concept.
First, let's begin by calculating the number of moles of B2H6 produced. To do this, we use the molar mass of B2H6.
The molar mass of B2H6 is:
(2 x atomic mass of B) + (6 x atomic mass of H)
= (2 x 10.81 g/mol) + (6 x 1.01 g/mol)
= 21.62 g/mol + 6.06 g/mol
= 27.68 g/mol
Next, we use the molar mass of B2H6 to calculate the number of moles produced:
Number of moles = mass / molar mass
Number of moles = 0.529 g / 27.68 g/mol
Number of moles ≈ 0.0191 mol
From the balanced chemical equation, we see that there is a 1:1 stoichiometric ratio between NaBH4 and B2H6. This means that for every 1 mole of NaBH4, we produce 1 mole of B2H6.
Now, we can use the molar concentration of NaBH4 to convert moles to volume:
Molarity = moles / volume
0.500 M = 0.0191 mol / volume
Rearranging the equation, we can solve for the volume:
Volume = moles / molarity
Volume = 0.0191 mol / 0.500 M
Volume ≈ 0.0382 L
Finally, to convert the volume from liters to milliliters:
Volume (in mL) = Volume (in L) x 1000
Volume (in mL) = 0.0382 L x 1000
Volume (in mL) ≈ 38.2 mL
Therefore, approximately 38.2 mL of a 0.500 M solution of NaBH4 is required to produce 0.529 g of B2H6.
mols B2H6 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols B2H6 to mols NaBH4.
Now convert mols NaBH4 to volume with
M = mols/L solution. You know M and mols, solve for L solution and convert to mL.