Sodium borohydride, NaBH₄, a substance used in the synthesis of many pharmaceutical agents, can be prepared by reaction of NaH with B₂H₆ according to the equation 2 NaH + B2H₆ → 2 NaBH₄

(a) How many grams of NaBH₄ can be prepared by reaction between 8.55 g of NaH and 6.75 g of B₂H₆?

Video transcript

Hey everyone. So for this problem, we need to find the mass of a L 203 that forms an aluminum and oxygen react were given our balanced chemical equation so we can jump right into this to find the limiting reactant, we'll do 128 .7g of aluminum. And we need to multiply it by aluminum's molar mass conversion. So we have one mole of aluminum Is about 27 g of aluminum. Then we need to multiply it by the multiple ratio between aluminum and L 203. So for every four moles of aluminum, we have two moles of a L 203, so two moles A L 203. We have four moles of aluminum. When you solve that out, you'll find we have about 2.38 moles Of a little 203 That can be formed from 128.7 g of of aluminum. So we're going to do the same thing with oxygen. We have 183.55 g of oxygen And again multiplied by oxygen's Mueller mass conversion. So for one mole of oxygen, we have about 32 g of oxygen. Again, we want to multiply it by our multiple ratio. So we have three moles of oxygen for two moles, ale 203. So two moles A L for three moles of oxygen. When you solve this out, you'll find we have 3.82 moles of a L 203 things are formed. So are limiting reactant here is going to be aluminum because only 2. moles of a L 203 reformed from that. So all we need to do now is to convert that to 2030.38 g. So we have 2.38 moles of a L 203 And we need to multiply it by its moles to g conversion. So for every 102g of a L 203, we have one mole Of A L 203. When you solve this out, we're left with 2 42.76 g of a L 203. This is going to be our answer to this problem. I hope this video helped you out and I'll see you in the next one.