Question

Antimony reacts with sulphur what percentage yeild for reaction in which 1.40 g os Sb2S3 is obtained from 1.73g of antimony

Answer 1

Answer:

1. Note the molar masses :

Sb = 121.76 g/mol

S = 32 g/mol

Sb2S3 = 339.53 g/mol

2. Find the moles of Sb

1.73g Sb x (1mol/121.76) = .014mol Sb

3. Use a mole ratio to find the moles of Sb2S3

.014mol Sb x (1mol Sb2S3 / 2mol Sb) = .007 mol Sb2S3

4. Convert the moles of Sb2S3 to grams

.007mol Sb2S3 x (339.52g/mol/1mol) = 2.377 grams = theoretical (predicted) yield

5. Calculate the percent yield

percent yield = 100 x (actual/predicted)

percent yield = 100 x (1.4/2.377) = 58.9%

Answer 2

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