How would you solve this?
6 Copper(II) oxide reacts with ammonia to produce copper metal, water vapour and nitrogen gas according to the equation:
CuO(s) + 2NH3(aq) → Cu(s) + 3H2O(g) + N2(g)
What amount of copper metal (in mol) will result from the reduction of
0.335 mol of copper(II) oxide by excess ammonia?
10 points for best answer!
6 Copper(II) oxide reacts with ammonia to produce copper metal, water vapour and nitrogen gas according to the equation:
CuO(s) + 2NH3(aq) → Cu(s) + 3H2O(g) + N2(g)
What amount of copper metal (in mol) will result from the reduction of
0.335 mol of copper(II) oxide by excess ammonia?
10 points for best answer!
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The balanced reaction is as follows:
3CuO(s) + 2NH3(aq) → 3Cu(s) + 3H2O(g) + N2(g)
From this reaction it is evident that
3 moles of CuO(s) yields 3 moles of Cu(s)
OR 1 mole of CuO(s) will yield 1 moles of Cu(s)
Thus 0.335 mole of CuO will give 0.335 moles of Cu
OR amount of Cu = 0.335 x 63.5 = 21.27 g
3CuO(s) + 2NH3(aq) → 3Cu(s) + 3H2O(g) + N2(g)
From this reaction it is evident that
3 moles of CuO(s) yields 3 moles of Cu(s)
OR 1 mole of CuO(s) will yield 1 moles of Cu(s)
Thus 0.335 mole of CuO will give 0.335 moles of Cu
OR amount of Cu = 0.335 x 63.5 = 21.27 g