Find the actual yield when the theatrical yield is 1.09g and percent yield is 90%.
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As mentioned above, phosphoric acid has 3 pKa values, and after 3 ionization it gives 3 types of ions at different pKa values:
H₃PO₄(aq) + H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq) pKₐ₁
<span>
</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq) pKₐ₂
HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq) pKₐ₃
At the highest pKa value (12.4) of phosphoric acid, the last OH group will lose its hydrogen. On the picture I attached, it is shown required protonated form of phosphoric acid before reaction whose pKa value is 12.4.
H₃PO₄(aq) + H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq) pKₐ₁
<span>
</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq) pKₐ₂
HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq) pKₐ₃
At the highest pKa value (12.4) of phosphoric acid, the last OH group will lose its hydrogen. On the picture I attached, it is shown required protonated form of phosphoric acid before reaction whose pKa value is 12.4.
5. 1.16 x moles moles of pennies would be required to equal the mass of the moon.
6. 12.86 moles of ethanol are in a 750 ml bottle of vodka.
Explanation:
5 .Data given:
mass of penny = 2.5 grams
atomic mass of penny = 62.93 grams/mole
moles present in mass of the moon given as = 7.3 x kg
number of moles =
number of moles =
0.039 moles of penny is present in 2.5 grams
0.039 moles of penny in 2.5 grams of it
so, x moles in 7.3 X grams
x = 1.16 x moles
so when the mass of the penny given is equal to the mass of moon, number of moles of penny present is 1.1 x .
6.
Given:
vodka = 40% ethanol
volume of vodka bottle = 750 ml
moles of ethanol =?
density of ethanol =0.79 g/ml
atomic mass of ethanol = 46.07 grams/mole
so, from the density of ethanol given we can calculate how much ethanol is present in the solution.
density =
density x volume = mass
0.79 x 750 = 592.5 grams
number of moles =
number of moles of ethanol =
= 12.86 moles of ethanol