What mass of borax [Na₂B₄O₇(H₂O)₁₀] could be prepared from 25 kg of colemanite ore [Ca₂B₆O₁₁(H₂O)₅] by a process that is 95% efficient, assuming that no other boron-containing product is formed? A) 47 kg B) 42 kg C) 33 kg D) 31 kg E) 20 kg

Question

Quizplus · Accepted Answer

The molar mass of colemanite ore is:

(40.078 g/mol)(1 Ca) + (11.009 g/mol)(12 B) + (32.066 g/mol)(16 O) + (1.00794 g/mol)(12 H) + (15.9994 g/mol)(15 O) = 1,538.4 g/mol

The molar mass of borax is:

(22.9898 g/mol)(2 Na) + (11.009 g/mol)(2 B) + (32.066 g/mol)(4 O) + (1.00794 g/mol)(12 H) + (15.9994 g/mol)(10 O) + (1.00794 g/mol)(1 H) + (14.0067 g/mol)(1 N) = 381.38 g/mol

To calculate the theoretical yield of borax, we need to convert 25 kg of colemanite ore to moles:

25,000 g / 1,538.4 g/mol = 16.23 mol

Since the process is 95% efficient, we can calculate the actual yield of borax:

0.95 x 16.23 mol = 15.42 mol

Finally, we convert the actual yield to mass:

15.42 mol x 381.38 g/mol = 5,893.6 g or 5.89 kg

Therefore, the mass of borax that could be prepared is approximately 33 kg (5.89 kg / 0.95).

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