Asked by Juor
Ammonia reacts with copper(ll)oxide to give nitrogen gas, copper and steam. Balance the equation. If 18.1g of ammonia is reacted with 90.4g of coope(ll)oxide,how many moles of each reacts?,and how many grams of nitrogen are formed?
Answers
Answered by
MathMate
Cu(II) oxide is CuO
so
3CuO+2NH3 → 3Cu+N2+3H2O
From 2NH3 and 3CuO, we conclude that
2 moles of NH3 reacts with 3 moles of CuO to form 1 mol of N2.
Find the molar masses of NH3 and CuO and determine which is the limiting reagent. Use stoichiometric ratios to calculate the number of moles reacted and the mass of N2 formed.
so
3CuO+2NH3 → 3Cu+N2+3H2O
From 2NH3 and 3CuO, we conclude that
2 moles of NH3 reacts with 3 moles of CuO to form 1 mol of N2.
Find the molar masses of NH3 and CuO and determine which is the limiting reagent. Use stoichiometric ratios to calculate the number of moles reacted and the mass of N2 formed.
Answered by
eugene
the limiting reactant is cuo
Answered by
motseki
0.757mole is reacted
7.059g of N2 is produced
7.059g of N2 is produced
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