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3 years ago

The combustion of 43.9g of ammonia produces ______ g of no2

1 answer:

Pie3 years ago

4 0

We calculate for the number of moles of ammonia using its molar mass:
moles of NH3 = 43.9g NH3 (1mol NH3/17.031g NH3)
= 2.578 mol
To compute for the number of moles of NO2, we use the mole ratio of ammonia and nitrogen dioxide from their coefficients in the balanced chemical equation of the combustion of NH3 in the presence of oxygen yielding nitrogen dioxide and water:
4NH3(g) + 7O2(g) = 4NO2(g) + 6H2O(g)
which is 4 moles of ammonia is to produce 4 moles of nitrogen dioxide:
moles NO2 = 2.578 mol NH3 (4mol NO2/4 mol NH3)
= 2.578 mol
Using the molar mass of NO2, the mass of NO2 produced is therefore
mass NO2 = 2.578 mol NO2(46.0055g NO2/1mol NO2)
= 118.6 grams NO2

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Mass of $1.4163\times 10^{-5}$ moles of red dye:

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Concentration of red solution after dilution = c'

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The final concentration of the diluted solution is $2.83\times 10^{-6} M$

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