A 50.0 g sample of liquid water at 0.0°c ends up as ice at -20.0°c. How much energy is involved in this change.
1 answer:
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Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles =
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O = = 0.13moles
Number of moles of NH₃ = = 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g
Answer:
<h2>D) 6</h2>Explanation:
since, n = molar mass / empirical formula mass
Empirical formula mass = Total mass of atoms present in empirical formula
CHCl = 12+1+35.5
= 48.5
Given, Molar mass = 290.8 g.
So, n = 290.8/48.5
= 5.995 , that is approx 6.
So, Molecular formula = n × Empirical formula
= 6 × CHCl
=
So, Number of C = 6