Answer: 61 grams
Explanation:
To calculate the number of moles, we use the equation:


The chemical equation for the combustion of octane in oxygen follows the equation:
By stoichiometry of the reaction;
25 moles of oxygen react with 2 moles of octane
4.69 moles of oxygen react with=
moles of octane
Thus, oxygen is the limiting reagent as it limits the formation of product and octane is the excess reagent.
25 moles of oxygen produce 18 moles of water
4.69 moles of oxygen produce=
moles of water.
Mass of water produced=
The maximum mass of water that could be produced by the chemical reaction is 61 grams.
It applies to both physical and chemical changes.
Answer:
4N2H4(2) + 2N204 (1) --> 6N2 (g) + 8H20 (g), AHO-2098 kJ
Explanation:
2N2H4(2) + N204 (1) --> 3N2 (g) + 4H20 (g),AHO-1049 kJ
when 6 mol of nitrogen are formed?
The ratio of Nitrogen in the previous therrmochemical equation to when 6 mol f nitrogen are formed is 2; 6/3 = 2.
So to get the new thermoochemical equation, multiply all parameters in the given equation by 2.
We have;
4N2H4(2) + 2N204 (1) --> 6N2 (g) + 8H20 (g), AHO-2098 kJ
<u>We are given:</u>
P1 = 3 atm T1 = 623 K <em>(350 + 273)</em>
P2 = x atm T2 = 523 K <em>(250 + 273)</em>
<em />
<u>Solving for x:</u>
From the idea gas equation:
PV = nRT
since number of moles (n) , Volume (V) and the Universal Gas constant(R) are constants;
P / T = k (where k is a constant)
the value of k will be the same for a gas with variable pressure and temperature and constant moles and volume
Hence, we can say that:
P1 / T1 = P2 / T2
3 / 623 = x / 523
x = 523 * 3 / 623
x = 2.5 atm (approx)
Therefore, the final pressure is 2.5 atm