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3 years ago

A sample of 8.5 g nh3 on oxidation produces 4.5 g of no. calculate the percent yield. reaction: 4 nh3 5 o2 → 4 no 6 h2o

1 answer:

5 0

30.% Determine the molar mass of NH3 and NO Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol Molar mass NO = 14.0067 + 15.999 = 30.0057 g/mol Moles NH3 = 8.5 / 17.03052 = 0.499103962 mol Moles NO = 4.5 / 30.0057 = 0.149971505 mol Looking at the balanced equation, for every mole of NH3 consumed, you should get one mole of NO. So if we had 100% yield, we should have 0.499103962 moles of NO. But we don't. The percent yield is a simple matter of division by what we did get by what we should get. So 0.149971505 / 0.499103962 = 0.300481497 = 30.0481497% Rounding to 2 significant figures gives 30.% yield.

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3 years ago

Read 2 more answers

How many grams of sodium benzoate, C H CO Na, have to be added to 1.50 L of a 0.0200 M solution of benzoic acid, C H CO H, to ma

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Answer: 2.8275grams

Explanation: A buffer is made btw a weak acid and it salt. In a solution made by dissolving a weak acid in solution, equilibrium is set up btw ionised and unionised ion. For Benzoic acid

C6H5COOH....> C6H5COO- + H+

Ka = [C6H5COO-] [H+]/ [C6H5COOH].......(1)

using Ka = 6.5× 10^-5, [C6H5COOH] = 0.02M. PH= - log[H+] ....> [H+]= 10^-4M.

Putting the values in(1)

[C6H5COO-]= 6.5× 10^-5 × 0.02/ 10^-4

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Mole(C6H5COONa) = 0.013 × Volume = 0.013mol/litre × 1.5 litre

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Mass(C6H5COONa) = 2.8275g

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Because both CaCl2 and CaBr2 both contain elements (Chlorine and Bromine) from the same group (group 7)

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