Answer:
10 moles of SO₂ are produced when 5 moles of FeS₂
Explanation:
Stoichiometry: it is the theoretical proportion in which the chemical species are combined in a chemical reaction. The stoichiometric equation of a chemical reaction relates molecules or number of moles of all the reagents and products that participate in the reaction.
In other words, stoichiometry establishes relationships between the molecules or elements that make up the reactants of a chemical equation with the products of said reaction. The relationships established are molar relationships (that is, moles) between the compounds or elements that make up the chemical equation.
The stoichiometric coefficients of a chemical reaction indicate the proportion in which said substances react.
Taking into account the above, you can apply the following rule of three: by stoichiometry if 4 moles of FeS₂ produce 8 moles of SO₂, then when reacting 5 moles of FeS₂ how many moles of SO₂ will they produce?
moles of SO₂= 10
<u><em>10 moles of SO₂ are produced when 5 moles of FeS₂</em></u>
Density(D) is defined as Mass(M) divided by Volume(V).
The formula for Density is:
D = M / V.
Another way to remember the formula for Density is to remember "Mass per unit of volume".
I hope this helps!
Answer:
0.054 mol O
Explanation:
<em>This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of carbon in a sample of acetic acid. How many moles of oxygen are in the sample?</em>
<em />
Step 1: Given data
- Chemical formula of acetic acid: CH₃CO₂H
- Moles of carbon in the sample: 0.054 moles
Step 2: Establish the appropriate molar ratio
According to the chemical formula, the molar ratio of C to O is 2:2.
Step 3: Calculate the moles of oxygen in the sample
We will use the molar ratio to determine the moles of oxygen accompanying 0.054 moles of carbon.
0.054 mol C × (2 mol O/2 mol C) = 0.054 mol O
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
