Answer:
–2733.4 KJ
Explanation:
The balanced equation for the reaction is given below:
C₂H₅OH + 3O₂ —> 2CO₂ + 3H₂O
ΔH = −1366.7 kJ
From the balanced equation above,
1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.
Finally, we shall determine the enthalpy change (ΔH) produced by the reaction of 2 moles of C₂H₅OH. This can be obtained as follow:
From the balanced equation above,
1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.
Therefore, 2 moles of C₂H₅OH will react to produce enthalpy change (ΔH) of = 2 × −1366.7 = –2733.4 KJ.
Thus, enthalpy change (ΔH) obtained is –2733.4 KJ
Answer:
A There are the same number of atoms of oxygen and hydrogen in the reactants as there are in the products.
Explanation:
Matter is conserved in a chemical equation if there are no loss of atoms. That means, total number of atoms of elements must be the same as the total number of element on the product side. The correct option is;
A. There are the same number of atoms of oxygen and hydrogen in the reactants as there are in the products.
Answer:
See explanation
Explanation:
The equation of the reaction is;
C3H8 + 5O2 ----> 3CO2 + 4H2O
Number of moles of C3H8 = 132.33g/44g/mol = 3 moles
1 mole of C3H8 yields 3 moles of CO2
3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2
Number of moles of oxygen = 384.00 g/32 g/mol = 12 moles
5 moles of oxygen yields 3 moles of CO2
12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2
Hence C3H8 is the limiting reactant.
Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2
1 moles of C3H8 yields 4 moles of water
3 moles of C3H8 yields 3 × 4/1 = 12 moles of water
Mass of water = 12 moles of water × 18 g/mol = 216 g of water
b) Actual yield = 269.34 g
Theoretical yield = 396 g
% yield = actual yield/theoretical yield × 100/1
% yield = 269.34 g /396 g × 100
% yield = 68%
Answer:
you are right
Explanation:
and i do not need to explain it because you did
Iam willing to play truth and dare
Explanation:
because I love to play truth and dare
