Sodium borohydride (NaBH_4) is used industrially in many organic syntheses. One

Question

Sodium borohydride (NaBH_4) is used industrially in many organic syntheses. One way to prepare it is by reacting sodium hydride (NaH) with gaseous diborane (B_2 H_6). A reaction was prepared by mixing 7.98 g of sodium hydride (NaH) with 8.26 g of gaseous diborane (B_2 H_6)? What is the balanced chemical reaction for sodium hydride (NaH) with gaseous diborane (B_2 H_6)? How many moles of NaH are in 7.98 g of sodium hydride? How many moles of B_2 H_6 are in 8.26 g of diborane? Which is the limiting reactant Why? What is the theoretical yield (in grams) of NaBH_4 that can be prepared by reacting 7.98 g of sodium hydride with 8.26 g of diborane? Assuming an 88.5% yield, how many grams of NaBH_4 can be prepared by reacting 7.98 g of sodium hydride with 8.26 g of diborane?

Explanation / Answer

The balanced equation is 2NaH + B2H6 2NaBH4. No. Of moles of NaH = weight / molecular mass = 7.98/23.998 = 0.3325 moles No. Of moles of B2H6 = 8.26/27.67 = 0.2985 moles The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. If all of the 0.2985 moles of B2H6 were to be used up, there would need to be 2 x 0.2985 = 0.597 moles of NaH. There is only 0.3325 moles of NaH available which makes it the limiting reactant. Hence NaH is the limiting reagent in this reaction. Limiting reagent is NaH. Hence 0.3325 moles produces 0.3325 moles of product. Molecular mass of NaBH4 is 37.83 g/mol                                                       Theoretical yield = Weight of NaBH4 = 37.83 x 0.3325 = 12.58 g Assuming 88.5 % yield, amount of NaBH4 formed = 12.58 x88.5 /100= 11.13 g

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