Answer:
10 molecules of NH₃.
Explanation:
N₂ + 3H₂ --> 2NH₃
As the N₂ supply is unlimited, what we need to do to solve this problem is <u>convert molecules of H₂ into molecules of NH₃</u>. To do so we use the <em>stoichiometric coefficients</em> of the balanced reaction:
10 NH₃ molecules could be prepared from 15 molecules of H₂ and unlimited N₂.
Answer:
The answers to the question are
(a) The the mg/L of H₃PO₄is 1462.05 g
(b) The molarity of H₃PO₄ in the solution is 14.92 mol/L
(c) The normality of H₃PO₄ in the solution 44.85 eq/L
Explanation:
(a) The commercially available H₃PO₄ has a concentration of 85.5 percent by weight, therefore
The mass of H₃PO₄ is found by
Density of water = 1.71 g/ml
mass of one liter of acid solution = 1.71 ×1000 = 1710 g
Therefore 85.5 % by weight of H₃PO₄ =1710×85.5/100 = 1462.05 g
Therefore we have 1462.05 g/L of H₃PO₄
Molar mass of H₃PO₄ = 97.994 g/mol
(b)Therefore the number of moles in 1462.05 g = 14.92 moles and the molarity = 14.92 mol/L
(c) The Normality = = 1465.05÷(1×97.994/3) = 44.85 eq/L