<h3>Answer:</h3>
0.64 Moles of Propane
<h3>Explanation:</h3>
Data:
Moles of Carbon = 1.5 mol
Conversion factor = 7 mol C produces = 3 mol of Propane
Solution:
As we know,
7 moles of Carbon produces = 3 moles of Propane
Then,
1.5 moles of Carbon will produce = X moles of Propane
Solving for X,
X = (1.5 moles × 3 moles) ÷ 7 moles
X = 0.6428571 moles of Propane
Or rounded to two significant figures,
X = 0.64 Moles of Propane
The red colour is the limiting reactant.
Red-blue colour ball and two white balls attached together are reactants.
Red-blue colour ball and two white and one red colour ball attached to each other are products.
<h3>What is a limiting reagent?</h3>
The reactant that is entirely used up in a reaction is called a limiting reagent.
A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products.
A product is a substance that is present at the end of a chemical reaction.
Hence,
The red colour is the limiting reactant.
Red-blue colour ball and two white balls attached together are reactants.
Red-blue colour ball and two white and one red colour ball attached to each other are products.
Learn more about limiting reagents here:
brainly.com/question/26905271
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Answer:
Explanation:
Molar ratio for Sg : FeS = 1:8
If there are 0.3 moles for Sg
Therefore, 0.3 × 8 =2.4 moles of FeS
Mass = Moles/ Mr
Mr of FeS = 56+32=88
So mass = 2.4/88
Mass= 0.027g
Answer:
63.53% yield
Explanation:
The balanced equation for this reaction is 2NaCl + H2O -> 2NaOH +Cl2
First we must find the limiting reactant
From NaCl we can only produce 6.06 grams of Cl2 in <u>theory</u>
From H20 we can only produce 38.995 grams in theory
so we know NaCl is the limiting
% yield is (Actual/Theoretical) x100 so
(3.85/6.06)x100= 63.53% yield
Answer:
(a)
(b)
Explanation:
Hello,
(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:
![\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3Dkt%20%2B%5Cfrac%7B1%7D%7B%5BNOBr%5D_0%7D%5C%5C%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3D%5Cfrac%7B0.8%7D%7BM%2As%7D%2A22s%2B%5Cfrac%7B1%7D%7B0.086M%7D%3D%5Cfrac%7B29.3%7D%7BM%7D%5C%5C)
![[NOBr]=\frac{1}{29.2/M}=0.0342M](https://tex.z-dn.net/?f=%5BNOBr%5D%3D%5Cfrac%7B1%7D%7B29.2%2FM%7D%3D0.0342M)
(b) Now, for a second-order reaction, the half-life is computed as shown below:
![t_{1/2}=\frac{1}{k[NOBr]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BNOBr%5D_0%7D)
Therefore, for the given initial concentrations one obtains:
Best regards.
