Hydrolysis of the compound B5H9 forms boric acid, H3BO3. Fusion of boric acid with sodium oxide forms a borate salt, Na2B4O7. Part A Without writing complete equations, find the mass (in grams) of B5H9 required to form 148 g of the borate salt by this reaction sequence.

Question

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Respuesta :

nuhulawal20 nuhulawal20 · Answer 1 · 2021-06-06T07:19:46+02:00

Answer:

37.09 g of B₅H₉ is required to form 148 g of borate salt( Na₂B₄O₇ )

Explanation:

Given the data in the question;

B₅H₉ → H₃BO₃ → Na₂B₄O₇

find the mass (in grams) of B5H9 required to form 148 g of the borate salt by this reaction sequence.

S M.W

B₅H₉ 63

H₃BO₃ 62

Na₂B₄O₇ 201

So, moles of Na₂B₄O₇ = 148 / 201 = 0.736

Since Boron will be conserved,

Moles of Boron atoms in Na₂B₄O₇ will be;

⇒ 4 × 0.736 = 2.944

Now, Moles of Boron in Na₂B₄O₇ = Moles of Boron in H₃BO₃ = Moles of Boron in B₅H₉

Hence,

Moles of Boron in B₅H₉ = 2.944

Moles of B₅H₉ = 2.944 / 5 = 0.5888

Mass of B₅H₉ = 0.5888 × 63 = 37.09 g

Therefore, 37.09 g of B₅H₉ is required to form 148 g of borate salt( Na₂B₄O₇ )