Answer:
11.3 g of H₂O will be produced.
Explanation:
The combustion is:
2C₈H₁₈ + 25O₂→ 16CO₂ + 18H₂O
First of all, we determine the moles of the reactants in order to find out the limiting reactant.
8 g / 114g/mol = 0.0701 moles of octane
37g / 32 g/mol = 1.15 moles of oxygen
The limiting reagent is the octane. Let's see it by this rule of three:
25 moles of oxygen react to 2 moles of octane so
1.15 moles of oxygen will react to ( 1.15 . 2)/ 25 = 0.092 moles of octane.
We do not have enough octane, we need 0.092 moles and we have 0.0701 moles. Now we work with the stoichiometry of the reaction so we make this rule of three:
2 moles of octane produce 18 moles of water
Then 0.0701 moles of octane may produce (0.0701 . 18)/2= 0.631 moles of water.
We convert the moles to mass → 0.631 mol . 18 g/1mol = 11.3 g of H₂O will be produced.
Nitrogen combine with hydrogen to produce ammonia 
at a 
ratio:
Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. 
of hydrogen gas would have been consumed while 
of ammonia would have been produced. The final mixture would therefore contain
Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:
Answer:
Li and H
Explanation:
2Li(s)+2H2O(i)→2LiOH(aq)+H2(g) is full balanced
It is called a polyatomic ion.
Hope this helps!!!
Answer: B:
Explanation:
This is the most reasonable answer
of
and