<u>Answer:</u> The given amount of iron reacts with 9.0 moles of
and produce 6.0 moles of 
<u>Explanation:</u>
We are given:
Moles of iron = 12.0 moles
The chemical equation for the rusting of iron follows:

By Stoichiometry of the reaction:
4 moles of iron reacts with 3 moles of oxygen gas
So, 12.0 moles of iron will react with =
of oxygen gas
- <u>For iron (III) oxide:</u>
By Stoichiometry of the reaction:
4 moles of iron produces 2 moles of iron (III) oxide
So, 12.0 moles of iron will produce =
of iron (III) oxide
Hence, the given amount of iron reacts with 9.0 moles of
and produce 6.0 moles of 
Answer:
0.208mole of CO2
Explanation:
First, let us calculate the number of mole of HC3H3O2 present.
Molarity of HC3H3O2 = 0.833 mol/L
Volume = 25 mL = 25/100 = 0.25L
Mole =?
Mole = Molarity x Volume
Mole = 0.833 x 0.25
Mole of HC3H3O2 = 0.208mole
Now, we can easily find the number of mole of CO2 produce by doing the following:
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
From the equation,
1mole of HC2H3O2 produced 1 mole of CO2.
Therefore, 0.208mole of HC2H3O2 will also produce 0.208mole of CO2
Answer:
falls to 5.00 M. He'll do this by adding distilled water to the solution until it reaches a certain final volume.Calc
Explanation:
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hey there!:
density = 75.0 g/mL
Volume = 12 mL
mass = ?
Therefore:
D = m / V
75.0 =m / 12
m = 75.0 * 12
m = 900 g
Answer B
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