Answer:
F₂ (g) + FeI₂ (aq) → FeF₂ (aq) + I₂ (l)
Explanation:
Our reactants are:
F₂ → Fluorine gas, a dyatomic molecule
FeI₂ → Iron (II) iodine
Our products are:
I₂ → Iodine
FeF₂ → Iron (II) fluoride
Then, the reaction is:
F₂ (g) + FeI₂ (aq) → FeF₂ (aq) + I₂ (l)
We see it is completely balanced.
Answer:
d. 12.3 grams of Al2O3
Explanation:
Based on the reaction:
4Al + 3O2 → 2Al2O3
<em>Where 4 moles of Al reacts in excess of oxygen to produce 2 moles of aluminium oxide.</em>
<em />
To solve this question we must find the moles of Aluminium. With these moles we can find the moles of aluminium oxide using the reaction:
<em>Moles Al -Molar mass: 26.9815g/mol-</em>
6.50g * (1mol / 26.9815g) = 0.241 moles Al
<em>Mass Al₂O₃ -Molar mass: 101.96g/mol-</em>
0.241 moles Al * (2 mol Al2O3 / 4 mol Al) = 0.120 moles Al2O3
0.120 moles Al2O3 * (101.96g / mol) =
12.3g of Al2O3 are produced.
Right answer is:
<h3>d. 12.3 grams of Al2O3
</h3>
Answer:
moles = given mass/atomic mass
so H2O mass = 2 +16=18
so 12g of h2o= 12/16 = 3/4 moles
Answer:
See Explanation
Explanation:
moles of NH₃ = 11.9g/17.03 g/mol = 0.699 mole
moles of CN₂OH₄ = 1/2(0.699) mole =0.349 mole
Theoretical yield of CN₂OH₄ = (0.349 mole)(60 g/mole) = 20.963 grams
%Yield = Actual Yield/Theoretical Yield x 100%
= 18.5g/20.963g x 100% = 88.25%
