Answer: Limiting reactant = 3
Theoretical Yield= 1
Excess reactant=2
Explanation: The theoretical yield is the maximum possible mass of a product that can be made in a chemical reaction. It can be calculated from: the balanced chemical equation. the mass and relative formula mass of the limiting reactant , and. the relative formula mass of the product.
An excess reactant is a reactant present in an amount in excess of that required to combine with all of the limiting reactant. It follows that an excess reactant is one remaining in the reaction mixture once all the limiting reactant is consumed.
The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. From the reaction stoichiometry, the exact amount of reactant needed to react with another element can be calculated
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
They're both a part of an atom.
2C3H8+ 702--->6CO2+8H20
FROM Equation above 2 moles of C3H8 reacted with 7 moles of oxygen to form 6 moles of c02 plus 8 molesof H2O
the moles of c3H8 reacted is = MASS/ R.F.M
THE R.F.M =48+8=44
Number of moles is hence 0.025/44=5.68x10^-4
since ratio of C3H8 to O2 is 2:7 Therefore moles of O2 reacted is 1.989 x10^-3
mass= r.f.m x number of moles
(1.989x10^-3) x 32 =0.064g
MAGNESIUM(Mg) posses the up mentioned electronic configuration
