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
Do you mean diborane? Not the usual method to make NaBH4!
2NaH + B2H6 -> 2NaBH4
work out the number of moles of each of the two starting materials so to find out if one is a limiting reagent. Then base your theoretical yield on the limiting reagent.
Once you have the theoretical yield then 0.868 x theoretical yield is the mass produced.
To calculate the theoretical yield of NaBH4, you need to determine the limiting reagent first. This can be done by comparing the stoichiometry of the reactants.
The balanced chemical equation for the reaction is:
4 NaH + B2H6 -> 2 NaBH4
Now, let's find the molar masses:
Molar mass of NaH = 23.0 g/mol
Molar mass of B2H6 = 27.67 g/mol
Next, calculate the number of moles of each reactant:
Moles of NaH = 7.98 g / 23.0 g/mol = 0.347 moles (rounded to 3 decimal places)
Moles of B2H6 = 8.16 g / 27.67 g/mol = 0.295 moles (rounded to 3 decimal places)
Based on the balanced equation, the stoichiometry is 4:1 for NaH to B2H6. Therefore, we have an excess of NaH, and it will be the limiting reagent.
Now, let's calculate the theoretical yield of NaBH4 using the limiting reagent:
Molar mass of NaBH4 = 37.83 g/mol
Moles of NaBH4 = moles of limiting reagent (NaH) * (2 NaBH4 / 4 NaH)
= 0.347 moles * (2 / 4)
= 0.1735 moles (rounded to 4 decimal places)
Finally, calculate the grams of NaBH4:
Grams of NaBH4 = moles of NaBH4 * molar mass of NaBH4
= 0.1735 moles * 37.83 g/mol
= 6.558 grams (rounded to 3 decimal places)
Therefore, the theoretical yield of NaBH4 is 6.558 grams. However, the given percentage yield is 86.8%, so we need to calculate the actual yield:
Actual yield = theoretical yield * percentage yield / 100
= 6.558 grams * 86.8 / 100
= 5.699 grams (rounded to 3 decimal places)
Hence, the actual yield of NaBH4 is 5.699 grams.
To determine the grams of NaBH4 that can be prepared, we need to calculate the limiting reagent and then use stoichiometry to find the yield. Here's how you can solve the problem step by step:
1. Write the balanced equation for the reaction:
4 NaH + 2 B2H6 -> 4 NaBH4
2. Calculate the number of moles for each reactant:
Moles of NaH = mass of NaH / molar mass of NaH
Moles of NaH = 7.98 g / (22.99 g/mol + 1.0079 g/mol) = 0.317 mol
Moles of B2H6 = mass of B2H6 / molar mass of B2H6
Moles of B2H6 = 8.16 g / (10.81 g/mol + 1.0079 g/mol) = 0.725 mol
3. Determine the limiting reagent:
The ratio between NaH and B2H6 in the balanced equation is 4:2, which simplifies to 2:1.
Since the molar ratio of NaH to B2H6 is 0.317 mol : 0.725 mol (approximately 0.437), NaH is the limiting reagent.
4. Calculate the moles of NaBH4 that can be formed based on the limiting reagent:
According to the balanced equation, 4 moles of NaBH4 are produced for every 4 moles of NaH.
Therefore, the moles of NaBH4 = 0.317 mol
5. Calculate the theoretical yield of NaBH4:
The theoretical yield is the maximum amount of product that can be obtained based on the limiting reagent.
The molar mass of NaBH4 is calculated as follows:
Molar mass of NaBH4 = molar mass of Na + molar mass of B + 4 * molar mass of H
Molar mass of NaBH4 = 22.99 g/mol + 10.81 g/mol + 4 * 1.0079 g/mol = 37.83 g/mol
The theoretical yield of NaBH4 = moles of NaBH4 × molar mass of NaBH4
Theoretical yield of NaBH4 = 0.317 mol × 37.83 g/mol = 12.0 g
6. Calculate the actual yield of NaBH4 (considering the 86.8% yield):
Actual yield = theoretical yield × yield percentage
Actual yield = 12.0 g × 0.868 = 10.41 g
Therefore, the actual yield of NaBH4 that can be prepared by reacting 7.98 g of sodium hydride and 8.16 g of diborone is approximately 10.41 grams.