I don't not know I did not learn this yet
Answer:
a) 4869 kj will be released
b) 43.86 g of octane
Explanation:
The heat of combustion is the amount of heat released when one mole of a substance reacts with enough oxygen
since the heat of combustion is per mol of combustible substancewjat we are required to do in this problem is calculate number moles in the reactions although in a different manner
a) MW C3H6O = 158 g/ mol
mol C3H6O = 158 g × 1 mol/ 58.08 g
= 2.72 - 1790 kj / mol ×2.72 mol = 4869 kj
b) Here we are asked the mass of octane to produce 1950 kj of heat knowing that per mole of octane we get 5074.1 kj then
1 mol / 5074.1kj × 1950 kj= 0.384 mol
mass C8H18 = 0.384 mol × 114.23 g/ mol = 43.86 g
Answer:
94.58 g of 
Explanation:
For this question we have to start with the reaction:
Now, we can balance the reaction:
We have the amount of 
and the amount of 
. Therefore we have to find the limiting reactive, for this, we have to follow a few steps.
1) Find the moles of each reactive, using the molar mass of each compound (
).
2) Divide by the coefficient of each compound in the balanced reaction ("2" for and "1" for ).
<u>Find the moles of each reactive</u>
<u>Divide by the coefficient</u>
<u />
The smallest values are for , so hydrogen is the limiting reagent. Now, we can do the calculation for the amount of water:
We have to remember that the molar ratio between and is 2:2 and the molar mass of is 18 g/mol.
There is 1 aluminium atom and 4 polonium atoms
