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# Annoying Equilibrium watch

1. Need help on this one

Nitrogen dioxide dissociates in the following equation

2NO2 <-------> 2NO + O2

21.3g of nitrogen dioxide were heated to a constant temperature in a flask of volume 11.5, an equilibrium mixture was formed which contained 7.04g of oxygen

a) Calc. the no of mole of oxygen present in the equilibrium mixture and deduce the no of moles of nitrogen monoxide present in the mixture.

for this i got 0.22 and 0.44 mol respectively

b) calc the no of moles in the original 21.3 g of nitrogen dioxide present and hence calc the no of moles present in the eqm mixture for nitrogen dioxide.

So the no. of moles of nitrogen dioxide present in 21.3 g are 0.463 mol, what will be the amount left in the equilibrium mixture( i notice the ratios are different but where from there?

many thanks
2. (Original post by devan5)
Need help on this one

Nitrogen dioxide dissociates in the following equation

2NO2 <-------> 2NO + O2

21.3g of nitrogen dioxide were heated to a constant temperature in a flask of volume 11.5, an equilibrium mixture was formed which contained 7.04g of oxygen

a) Calc. the no of mole of oxygen present in the equilibrium mixture and deduce the no of moles of nitrogen monoxide present in the mixture.

for this i got 0.22 and 0.44 mol respectively

b) calc the no of moles in the original 21.3 g of nitrogen dioxide present and hence calc the no of moles present in the eqm mixture for nitrogen dioxide.

So the no. of moles of nitrogen dioxide present in 21.3 g are 0.463 mol, what will be the amount left in the equilibrium mixture( i notice the ratios are different but where from there?

many thanks
From the stoichiometry of the equation the number of moles of NO formed equals the numberof moles of NO2 used up.

So the number of moles NO2 at equilibrium = moles initially - moles used up.
3. (Original post by charco)
From the stoichiometry of the equation the number of moles of NO formed equals the numberof moles of NO2 used up.

So the number of moles NO2 at equilibrium = moles initially - moles used up.
so its 0.463-0.44

where there is 0.218 mol of HI in the sample at the start and in the equilibrium mixture there is 0.023 mol of Hydrogen.

I understand that there will be also 0.023 mol of Iodine but what will be the no.of mol of HI in the eqm mixture, using the fact that there there is
a ratio.
4. (Original post by devan5)
so its 0.463-0.44

where there is 0.218 mol of HI in the sample at the start and in the equilibrium mixture there is 0.023 mol of Hydrogen.

I understand that there will be also 0.023 mol of Iodine but what will be the no.of mol of HI in the eqm mixture, using the fact that there there is
a ratio.
Once again you can se from the equation that dissociation of 2 moles of HI produces 1 mole of each of the products.

So initial moles of HI = 0.218 mol

Moles HI dissociated = 0.023 x 2 = 0.046 mol

Moles HI remaining = 0.218 - 0.046 mol

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