Answer:
It can't be done.
Explanation:
If you have only 5.4 g of oxygen, the most lithium oxide you can get is 7.7 g.
Only 2.3 g of lithium will react. and the other 22.3 g of lithium will not be used.
Answer:
Kc = 1.09x10⁻⁴
Explanation:
<em>HF = 1.62g</em>
<em>H₂O = 516g</em>
<em>F⁻ = 0.163g</em>
<em>H₃O⁺ = 0.110g</em>
<em />
To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:
Kc = [H₃O⁺] [F⁻] / [HF]
<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>
<em />
[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M
[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M
[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M
Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]
<h3>Kc = 1.09x10⁻⁴</h3>
Answer:
% yield = 82.5%
Explanation:
HgO + 2Cl₂ → HgCl₂ + Cl₂O
Our reactants are:
Our products are:
We do not have information about moles of reactants, but we do know the theoretical yield and the grams of product, in this case Cl₂O, we have produced.
Percent yield = (Yield produced / Theoretical yield) . 100
Theoretical yield is the mass of product which is produced by sufficent reactant. We replace data:
% yield = (0.71 g/0.86g) . 100 = 82.5%
