Answer:
1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
2) The amount (in grams) of excess reactant H₂ = 4.39 g.
Explanation:
- Firstly, we should write the balanced equation of the reaction:
<em>N₂ + 3H₂ → 2NH₃.</em>
<em>1) To determine the limiting reactant of the reaction:</em>
- From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- This means that <em>N₂ reacts with H₂ with a ratio of (1:3).</em>
- We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation:<em> n = mass / molar mass.</em>
The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.
The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.
- From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).
The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).
∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>
- As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
- Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
- The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.
- ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.
Answer:
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Answer:
69.7% is percent yield
Explanation:
Based on the reaction:
3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)
2 moles of Na3PO4 react producing 6 moles of NaNO3.
As 24.2 moles of Na3PO4 react, theoretical moles of NaNO3 produced are:
24.2 moles Na3PO4 * (6 moles NaNO3 / 2 moles Na3PO4) =
72.6 moles of NaNO3
As there are produced 50.6 moles of NaNO3, percent yield is:
50.6 moles NaNO3 / 72.6 moles NaNO3 =
<h3>69.7% is percent yield</h3>
Answer: A pair of elements will most likely form an ionic bond if one is a metal and one is a nonmetal. These types of ionic compounds are composed of monatomic cations and anions.
Explanation:
A pair of elements will most likely form an ionic bond if one is a metal and one is a nonmetal. These types of ionic compounds are composed of monatomic cations and anions.
Explanation:
The chart below shows monatomic ions formed when an atom loses or gains one or more electrons, and the ionic compounds they form. You can check your periodic table to see that the cations are monatomic ions formed from metals, and the anions are monatomic ions formed from nonmetals.
Each covalent H bond is nonpolar.
