All categories

3 years ago

What is minimum volume of oxygen required to react with 42.5 g of aluminum in the synthesis of aluminum oxide at stp?

1 answer:

6 0

26.833 liters

Aluminum oxide has a formula of Al₂O₃, which means for every mole of aluminum used, 1.5 moles of oxygen is required (3/2 = 1.5).

Given 42.5 g of aluminum divided by its atomic mass (26.9815385) gives 1.575 moles of aluminum.

Since it takes 1.5 moles of oxygen per mole of aluminum to make aluminum oxide, you'll need 2.363 moles of oxygen atoms.

Each molecule of oxygen gas has 2 oxygen atoms, so the moles of oxygen gas will be 2.363/2 = 1.1815

Finally, you need to calculate the volume of 1.1815 moles of oxygen gas.
1 mole of gas at STP occupies 22.7 liters of volume. Therefore,

1.1815 * 22.7 = 26.8 liters of oxygen gas.

You might be interested in

Help me with this question please.

Salsk061 [2.6K]

Explanation:

$current = \frac{voltage}{resistance} \\ i = \frac{v}{ohm} \\ i = \frac{240}{38} \\ i = 6.3158 \: amps$

hope it helped

pls mark BRAINLIEST

thanks

8 0

3 years ago

Convert 685K to Fahrenheit

joja [24]

773.33 degrees F
F=Fahrenheit
K=Kelvin
F=K x 9/5 - 459.67
773.33 = 685 x 9/5 - 459.67

8 0

3 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

3 years ago

Does mercury have more protons and electrons than tin?

Eva8 [605]

Yes. Mercury has 80 protons. Tin has 50 protons. Same for electrons, it just doesn't have an exact number.

6 0

3 years ago

At an ACME plant in Scottsdale Arizonia methanol is produced commercially

patriot [66]

Answer:

thats nice to know i will send Wile E. Coyote there for his supplys

Explanation:

4 0

3 years ago