How many grams of nitrogen dioxide may theoretically be produced when copper reacts completely with 21g of nitric acid?
Cu + 4HNO3 ---> Cu(NO3)2 + 2H2O + 2NO2
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It says the answer is 7.7 g. I don't know how to get it though. Please guide me through.
see 4 moles of HNO3 reacting with Cu produces 2 moles of NO2
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can you tell molecular mass of HNO3 and NO2
63 and 46
So there is 1/3 mol of HN03
now i convert those moles to gram
using equation moles * molar mass =gram
that is 4*63 gram of HNO3 will give rise to 2*46 g of NO2
then 21 gram of HNO3 gives = 21*2*46 /4*63 = 7.66
that is 4*63g/mol gives rise to 2*46g/mol right?
Yaaa that also u can doo :) since they have asked in grams i have initially only converted to gram u may also convert moles to gram also :)
So basically, here we have 4 mol * 63g/mol gives rise to 2 mol * 46g/mol.
The mols cancel out and u see that 252 g give rise to 92 g of NO2.
Since there is only 21 g of HNO3 then 252/x = 21 implies x = 12. Dividing 92/12 = 7.66666.