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Volume calculation


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#1 Icemaiden

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Posted 30 May 2005 - 07:32 PM

4NH3(g) + 5O2 --> 4NO(g) + 6H2O(l)

400L of oxygen is used up in the above reaction, measured under conditions in which its density is 1.6 g l-1.What mass of ammonia must have reacted?

I got 1.3kg but the book says 272g

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#2 dfx

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Posted 30 May 2005 - 07:40 PM

The density is 1.6 g/l . So 400 l is thus 460g.

460g of Oxygen is 20 moles (460/32 = 20 using n = m/gfm)

From the reaction, 4 moles NH3 reacts with 5 moles O2

Thus 16 moles of NH3 will react with 20 moles of O2

1 mole of NH3 is 17g

But 16 moles are reacting, somultiplied by 16 moles = 272g

Hope that helped smile.gif

#3 Icemaiden

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Posted 30 May 2005 - 07:47 PM

Thank-you....I think i went a bit wrong at the bottom in the mole ratios ..oops

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