Answer:
152.4 g of O₂ are consumed.
Explanation:
We start from the combustion reaction:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
We convert the mass of propane to moles:
41.9 g . 1mol /44g = 0.952 moles
Ratio is 1:5. 1 mol of propane consumes 5 moles of oxygen at propane combustion
Then, 0.952 moles may consume (0.952 . 5) /1 = 4.76 moles.
We convert moles to mass → 4.76 mol . 32g/mol = 152.4 g
So since every 20 mins the mass will be cut in half (and that happens 3 times), you have just got to divide 1g by 2, 3 times.
after 20 mins the mass will be .5
after 40 mins the mass will be .25
Answer: After 1 hour the mass will be .125g
Answer:
A)
.
B)
.
C) 0.9 mol.
D) Increasing both temperature and pressure.
Explanation:
Hello,
In this case, given the information, we proceed as follows:
A)
![N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)](https://tex.z-dn.net/?f=N_2%28g%29%2B3H_2%28g%29%5Crightleftharpoons%202NH_3%28g%29)
B) For the calculation of Kc, we rate the equilibrium expression:
![Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent
, we have:
![[NH_3]=0.6M=2*x](https://tex.z-dn.net/?f=%5BNH_3%5D%3D0.6M%3D2%2Ax)
![x=\frac{0.6M}{2}=0.3M](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B0.6M%7D%7B2%7D%3D0.3M)
Next, the concentrations of nitrogen and hydrogen at equilibrium are:
![[N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M](https://tex.z-dn.net/?f=%5BN_2%5D%3D%5Cfrac%7B1.5mol%7D%7B1.5L%7D-x%3D1M-0.3M%3D0.7M)
![[H_2]=\frac{4mol}{1.5L}-3*x=2.67M-0.9M=1.77M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%5Cfrac%7B4mol%7D%7B1.5L%7D-3%2Ax%3D2.67M-0.9M%3D1.77M)
Therefore, the equilibrium constant is:
![Kc=\frac{(0.6M)^2}{(0.7M)*(1.77M)^3}\\ \\Kc=0.0933](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%280.6M%29%5E2%7D%7B%280.7M%29%2A%281.77M%29%5E3%7D%5C%5C%20%5C%5CKc%3D0.0933)
C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.
D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.
Best regards.
<span>The number of molecules of each gas must be equal since 1 mol of any gas needs to contain 6.02x10^23 molecules. So 1mol O</span>₂ and 1mol CO both need to contain 6.02x10^23 molecules
Answer:
to mi understanding iz fusion reaction