Find the percent composition for Carbon Dioxide (CO2)
1 answer:
You might be interested in
Answer:
20 molecules of oxygen gas remains after the reaction.
Explanation:
Molecules of ethyne = 52
Molecules of oxygen gas = 150
According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.
Then 52 molecules of ethyne will react with:
of oxygen gas.
As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.
Remaining molecules of recessive reagent = 150 - 130 = 20
20 molecules of oxygen gas remains after the reaction.
Answer:
=> 1366.120 g/mL.
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 25 Kg
Volume (v) = 18.3 mL.
From our question, we are to determine the density (rho) of the rock.
The formula:
First let's convert 25 Kg to g;
1 Kg = 1000 g
25 Kg = ?
= 25000 g
Substitute the values into the formula:
= 1366.120 g/mL.
Therefore, the density (rho) of the rock is 1366.120 g/mL.