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<div>1. In an experiment 6.67 g of AlCl3 was produced and 0.54 g of Al remained unreacted.How many moles of Al were taken originally?</div> <div>&nbsp;</div> <div>&nbsp;</div> <div>&nbsp;</div> <div>2.One mole of Potassium is themally decomposed and excess of aluminium is burnt in the gaseous product.How many moles of aluminium oxide are formed?z</div> <div>&nbsp;</div> <div>&nbsp;</div> <div>3.A sample of H2SO4 is 90% by weight.What is the volume of the acid that can be used to &nbsp;make 1 litre of 0.2 M H2SO4</div> <div>&nbsp;</div> <div>4.The density of 3M solution of Na2S2O3 is 1.2 g/ml.Find the molalities of Na+ and S2O3^2- ions.</div> <div>&nbsp;</div> <div>5.If x% of a solute of molecular mass is present in a solution having density of d gm/cm^3 calculate molarity and molality if</div> <div>a. x% is by mass &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; b.x% is by volume</div>
Asked by mehrarora26 | 22 May, 2015, 09:43: PM
Dear mehrarora26@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.

In case of multiple questions within a query, please post each question individually and let us know where you are getting stuck so that we would be able to explain things better.

Al + 3/2 Cl2 ----> AlCl3
From the above reaction, 1mole Al produces 1 mole AlCl3
Finally 0.54 g Al is remaining that means 0.54 / 27 = 0.02 moles Al is remaining
6.67 g of AlCl3 is produced  that means 6.67 / 133.5 = 0.05 moles of AlCl3 is produced.
From the reaction, 0.05 moles of AlCl3 can get produced from 0.05 moles of Al.
Hence, 0.05 moles used Al + 0.02 moles remaining Al = 0.07 moles of Al were present originally.

Regards

Topperlearning Team.
Answered by Prachi Sawant | 23 May, 2015, 08:53: PM

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