References Use the References to access important values if needed for this question A student ran the following reaction in the
1 answer:
Answer:
19.27
Explanation:
Some values are corrected from correct source.Thus,
Moles of SO₂ = 8.19x10⁻² moles
Moles of O₂ = 8.10x10⁻² moles
Volume = 1 L
Concentration of SO₂ = 8.19x10² M
Concentration of O₂ = 8.10x10 M
Considering the ICE table for the equilibrium as:
2SO₂ (g) + O₂ (g) ⇔ 2SO₃ (g)
t = o 8.19x10⁻² 8.19x10⁻²
t = eq -2x -x 2x
--------------------------------------------- --------------------------
neq: 8.19x10⁻² -2x 8.19x10⁻² -x 2x
Given:
Equilibrium concentration of O₂ = 5.98x10⁻² M = 8.19x10⁻² -x
Thus, x = 0.0212 M
[SO₂] = 8.19x10⁻² - 2*0.0212 = 0.0395 M
[SO₃] = 2*0.0212 = 0.0424 M
The expression for the equilibrium constant is:
K = 19.27
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Answer:
127 grams of carbon dioxide
Explanation:
We need to determine the chemical equation first. Butane has a chemical formula of , oxygen is
, carbon dioxide is
, and water is
. The reactants are butane and oxygen and the products are carbon dioxide and water. So we write:
⇒
But remember! We need to balance this. Currently, there are 4 carbon atoms (C), 10 hydrogen atoms (H), and 2 oxygen atoms (O) on the left, while there are 1 carbon atom (C), 2 hydrogen atoms (H), and 3 oxygen atoms (O) on the right. Let's place a coefficient of 4 in front of the carbon dioxide and a coefficient of 5 on the water, so that we have equal numbers of carbon and hydrogen atoms on each side:
⇒
However, we need to ensure that there are equal numbers of O atoms, as well. On the left, we have 2 and on the right we have 13, so let's put a coefficient of 6.5 on the oxygen:
⇒
Finally, multiply everything by 2 to get whole number coefficients:
⇒
Ah, now we can actually get to the problem!
We need to determine the limiting reactant, so let's convert the 42 g of butane and 150 g of oxygen into moles of any product, say, carbon dioxide. To convert to moles, we need to find the molar mass of each compound.
The molar mass of butane is 4 * 12.01 + 10 * 1.01 = 58.14 g/mol, while the molar mass of oxygen is 2 * 16 = 32 g/mol. We can now set up the equations:
Clearly, we see that 2.8846 < 2.8896, which means that oxygen is the limiting reactant. In other words, the most products can be made when the oxygen is all used up.
Now let's finally convert moles of carbon dioxide into grams by multiplying by its molar mass, which is 12.01 + 2 * 16 = 44.01 g/mol:
Notice that we have 3 significant figures because we had 3 significant figures at the start with 150. grams of oxygen.
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