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Hatshy [7]
2 years ago
7

How many grams of Fe2O3 are formed when 51.3 grams of iron, Fe, react completely with oxygen, O2? 4 Fe + 3 O2 --> 2 Fe2O3

Chemistry
1 answer:
katovenus [111]2 years ago
6 0

The Fe is our limiting reactant here. First, convert the mass of Fe given to moles of Fe by dividing mass by the molar mass:

(51.3 g Fe)/(55.845 g/mol) = 0.9186 mol Fe.

According to the balanced equation, 2 moles of Fe2O3 are produced for every 4 moles of Fe that are consumed. In other words, half as many moles of Fe2O3 will be produced as there are Fe that react:

(0.9186 mol Fe)(2 mol Fe2O3/4 mol Fe) = 0.4593 mol Fe2O3.

Finally, convert the moles of Fe2O3 back to grams by multiplying by the molar mass of Fe2O3:

(0.4593 mol Fe2O3)(159.6882 g/mol) = 73.3 g Fe2O3.

So, 73.3 grams of Fe2O3 are formed.

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Answer:

We could do two 1:50 dilutions and one 1:4 dilutions.

Explanation:

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A solution that is 1000 ug/ ml  (or 1000 mg / l) is 1000 ppm.

Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.

Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.

We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):

First step (1:50 dilution):

Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).

Step 2 (1:50 dilution):

Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)

Step 3 (1:4 dilution):

Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.

6 0
3 years ago
What is the molarity of a KF(aq) solution containing 3.0 mol of KF in 2.00L of solution?
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Explanation : Given,

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Formula used :

\text{Molarity}=\frac{\text{Moles of }KF}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{3.0mol}{2.00L}=1.5mole/L=1.5M

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56g of Fe contains 6.02x10^23 atoms.

Therefore, 26.8g of Fe will contain = (26.8x6.02x10^23) / 56 = 2.881x10^23 atoms

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