sodium borohydride (NaBH4) is used industrially in many organic syntheses. one way to prepare it is by reacting sodium hydride with gaseous diborone. Assuming an 86.8% yield,how many grams of NaBH4 can be prepared by reacting 7.98g of sodium hydride and 8.16g of diborone?
i tried 9.68 but it didn't work
To determine the amount of NaBH4 that can be prepared, we need to calculate the theoretical yield based on the given reactant weights and the given percent yield.
Step 1: Find the molar mass of each compound.
- Sodium hydride (NaH): 23.00 g/mol for sodium + 1.01 g/mol for hydrogen = 24.01 g/mol
- Diborone (B2H6): 10.81 g/mol for boron + 1.01 g/mol for hydrogen = 11.82 g/mol
Step 2: Convert the given weights of NaH and B2H6 to moles.
- Moles of NaH = 7.98 g / 24.01 g/mol = 0.332 mol
- Moles of B2H6 = 8.16 g / 11.82 g/mol = 0.692 mol
Step 3: Determine the limiting reactant.
- The ratio of NaH to B2H6 is 1:1 based on the balanced equation.
- Since the mole ratio between NaH and B2H6 is 1:1, NaH is the limiting reactant because it has a smaller amount in moles.
Step 4: Calculate the theoretical yield of NaBH4.
- The molar mass of NaBH4 is 37.83 g/mol (22.99 g/mol for sodium + 1.01 g/mol for hydrogen + 10.81 g/mol for boron + 1.01 g/mol for hydrogen + 2.00 g/mol for hydrogen).
The theoretical yield can be calculated using the mole ratio between NaH and NaBH4:
- Moles of NaBH4 = Moles of NaH (0.332 mol) × (1 mol NaBH4 / 1 mol NaH) = 0.332 mol
- Theoretical yield of NaBH4 = Moles of NaBH4 × Molar mass of NaBH4 = 0.332 mol × 37.83 g/mol = 12.56 g
Step 5: Calculate the actual yield (assuming 86.8% yield).
- Actual yield of NaBH4 = Theoretical yield of NaBH4 × Percent yield = 12.56 g × 0.868 = 10.89 g
Therefore, the correct answer is 10.89 grams of NaBH4 can be prepared.
To solve this problem, we need to determine the limiting reactant and calculate the amount of NaBH4 produced based on the balanced chemical equation.
First, let's write the balanced chemical equation for the reaction:
3 NaH + B2H6 -> 2 NaBH4
According to the equation, for every 3 moles of NaH, we need 1 mole of B2H6 to produce 2 moles of NaBH4.
Step 1: Convert the given masses to moles.
Molar mass of NaH = 22.99 g/mol
7.98 g NaH / 22.99 g/mol = 0.347 mol NaH
Molar mass of B2H6 = 27.67 g/mol
8.16 g B2H6 / 27.67 g/mol = 0.295 mol B2H6
Step 2: Calculate the moles of NaBH4 produced from NaH and B2H6.
From the balanced equation, we know that 3 moles of NaH produce 2 moles of NaBH4.
So, the moles of NaBH4 produced from NaH = (0.347 mol NaH) * (2 mol NaBH4 / 3 mol NaH) = 0.231 mol NaBH4
Similarly, from the balanced equation, we know that 1 mole of B2H6 produces 2 moles of NaBH4.
So, the moles of NaBH4 produced from B2H6 = (0.295 mol B2H6) * (2 mol NaBH4 / 1 mol B2H6) = 0.590 mol NaBH4
Step 3: Determine the limiting reactant.
The limiting reactant is the one that produces the least amount of product. We compare the moles of NaBH4 produced from NaH and B2H6, and the reactant that produces fewer moles of NaBH4 is the limiting reactant.
In this case, NaH produces 0.231 mol NaBH4 and B2H6 produces 0.590 mol NaBH4. Thus, NaH is the limiting reactant.
Step 4: Calculate the mass of NaBH4 produced.
To find the mass of NaBH4 produced, we multiply the moles of NaBH4 produced from NaH by the molar mass of NaBH4.
Molar mass of NaBH4 = 37.83 g/mol
Mass of NaBH4 produced from NaH = 0.231 mol NaBH4 * 37.83 g/mol = 8.740 g NaBH4
Therefore, based on the given information and assuming an 86.8% yield, 8.740 grams of NaBH4 can be prepared by reacting 7.98 grams of sodium hydride and 8.16 grams of diborone.