150 grams of water because if 225 grams of propane are burned with 125 grams of oxyegen then there will be 150 grams of water
Answer:
The correct answer is 32.2 grams.
Explanation:
Based on the given information, the enthalpy of formation for aluminum oxide is 1676 kJ/mol. It signifies towards the energy that is required to generate aluminum and oxygen, and both of these exhibit zero enthalpy of formation. Therefore, the ΔHreaction is the required energy to generate 2 moles of aluminum. Thus, the energy needed for the formation of single mole of aluminum is,
ΔHrxn = 1676/2 = 838 kJ/mol
Q or the energy input mentioned in the given case is 1000 kJ. Therefore, the number of moles of Al generated is,
(1000 kJ) / (838 kJ/Al mole) = 1.19 moles of Aluminum
The grams of aluminum produced can be obtained by using the formula,
mass = moles * molecular mass
= 1.19 * 26.98
= 32.2 grams.
Answer:
6H2 + P4→ 4PH3
Explanation:
Phosphorus has 4 in it and hydrogen has 3 in it. in order to balance it, we have to put 4 in front of phosphine so that the phosphorus on the product side has an equal amount as to the one on the reactant side.
the only one left to balance is hydrogen and so in order to balance it we put a 6 on h2 because the hydrogen in the product size becomes 12 (4 * 3).
therefore the hydrogen on the reactant side becomes 12 as well (6 * 2)
Answer:
The answer to the question is
The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits
Explanation:
To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure
At the first trial the mixture contains
3.6 atm CO
1.2 atm H₂O (g)
Total pressure = 3.6+1.2= 4.8 atm
which gives
3.36 atm CO
0.96 atm H₂O (g)
0.24 atm H₂ (g)
That is
CO+H₂O→CO(g)+H₂ (g)
therefore the mixture contained
0.24 atm CO₂ and the total pressure =
3.36+0.96+0.24+0.24 = 4.8 atm
when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂
At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857
adding 1.8 atm CO gives 4.46 atm hence we have
(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857
which gives x = 0.031 atm or x = -0.6183 atm
Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm
