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d1i1m1o1n [39]
2 years ago
12

Use the balanced equation from number four above to determine the limiting reactant if we had a mixture of 5.0 moles of Fe and 4

.0 moles of O2.
Chemistry
1 answer:
Tom [10]2 years ago
8 0

Answer:

2Fe + 3O₂  →  2Fe₂O₃

Limiting reactant: O₂

2Fe + O₂ → 2FeO

Limiting reactant: Fe

Explanation:

2Fe + 3O₂  →  2Fe₂O₃

2Fe + O₂ → 2FeO

These are the possible reactions:

In first case, 2 moles of Fe need 3 mol of oyxgen to react

If I have 5 moles of Fe, I will need (5 .3)/2 = 7.5 moles of O₂

Then, the oxygen is my limiting ( I only have 4 moles)

3 moles of O₂ need 2 moles of Fe

If I have 4 moles of O₂, I will need (4 .2)/3 = 2.66 moles (I have 5)

Fe, is the reactant in excess.

For second case, 2 moles of Fe need 1 mol of O₂, to react.

If I have 5 moles of Fe, I will need ( 5 .1) / 2 =2.5 moles of O₂

I have 4 moles of oxygen, so now it is my excess.

1 mol of O₂ need 2 moles of Fe, to react

If I have 4 moles of O₂, I will need the double of amount, 8 moles of Fe.

I have 5 moles, then the Fe is my limtiing reactant.

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Answer:

131.8 g.

Explanation:

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<u><em>d = m / V.</em></u>

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2 years ago
During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

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it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

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using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

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