Answer:
4.5 moles of lithium sulfate are produced.
Explanation:
Given data:
Number of moles of lead sulfate = 2.25 mol
Number of moles of lithium nitrate = 9.62 mol
Number of moles of lithium sulfate = ?
Solution:
Chemical equation:
Pb(SO₄)₂ + 4LiNO₃ → Pb(NO₃)₄ + 2Li₂SO₄
Now we will compare the moles of lithium sulfate with lead sulfate and lithium nitrate.
Pb(SO₄)₂ : Li₂SO₄
1 : 2
2.25 : 2/1×2.25 = 4.5 mol
LiNO₃ : Li₂SO₄
4 : 2
9.62 : 2/4×9.62 = 4.81 mol
Pb(SO₄)₂ produces less number of moles of Li₂SO₄ thus it will act as limiting reactant and limit the yield of Li₂SO₄.
Answer:
Mass of NH₃ produced = 34 g
Explanation:
Given data:
Mass of nitrogen = 28 g
Mass of Hydrogen = 12 g
Mass of NH₃ produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 28 g/ 28 g/mol
Number of moles = 1 mol
Moles of hydrogen:
Number of moles = mass/molar mass
Number of moles = 12 g/ 2 g/mol
Number of moles = 6 mol
Now we will compare the moles of hydrogen and nitrogen with ammonia.
H₂ : NH₃
3 : 2
6 : 2/3×6 = 4 mol
N₂ : NH₃
1 : 2
Number of moles of ammonia produced by nitrogen are less thus it will act as limiting reactant.
Mass of ammonia produced:
Mass = number of moles × molar mass
Mass = 2 mol × 17 g/mol
Mass = 34 g
Answer:
Here's what I get.
Explanation:
- If your teachers don't ask for a specific type of formula, a condensed structural formula should be OK.
- If they ask specifically for a structural formula or a bond-line formula, that is what you must give.
Bottom line: ask your teachers in advance what they expect.
60 g C2H6 × 1 mol C2H6 x 7 mol O2 x 32 g O2 = ~223.5 g O2
30.068 g 2 mol C2H6 1 mol O2
