The balanced equation for the reaction is as follows;
Li₂O + H₂O ---> 2LiOH
Stoichiometry of Li₂O to H₂O is 1:1
Mass of Li₂O reacted - 18.9 g
Number of Li₂O moles reacted - 18.9 g / 30 g/mol = 0.63 mol
An equivalent amount of moles of water have reacted - 0.63 mol
mass of water required - 0.63 mol x 18 g/mol = 11.34 g
A mass of 11.34 g of water is required
Answer:
ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.
Explanation:
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:
products
Explanation:
yeah the answer is product