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3 years ago

1.10 atoms of aluminum reacts with 6 molecules of oxygen gas to produce aluminum oxide.

Chemistry

1 answer:

Elanso [62]3 years ago

3 0

Answer: Option A) 2 atoms

Explanation:

4Al + 302 ---> 2Al2O3

4 moles of Al reacts with 3 molecules of O2

So 1.10 moles of Al should reacts with Z molecules of O2 (Assume Z is unknown)

i.e 4 moles of Al = 3 molecules of O2

4 moles of Al = 6 atoms of O2 (since 1 molecule of oxygen has two atoms of oxygen)

1.10 atoms of Al = Z atoms of O2

To get the value of Z, cross multiply

Z x 4 moles = 6 atoms x 1.10 moles

4Z = 6.6

Divide both sides by 4

4Z/4 = 6.6/4

Z = 1.65 atoms (approximately 2 atoms)

Thus, 2 atoms of the excess reactant oxygen remains.

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Answer:

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3 years ago

Solid potassium chlorate decomposes into solid potassium chloride and oxygen gas. If 2.00 g potassium chlorate decomposes in a c

abruzzese [7]

Answer : The final chamber pressure is 0.746 atm.

Explanation:

First we have to calculate the moles of $KClO_3$ .

Molar mass of $KClO_3$ = 122.5 g/mole

$\text{ Moles of }KClO_3=\frac{\text{ Mass of }KClO_3}{\text{ Molar mass of }KClO_3}=\frac{2.00g}{122.5g/mole}=0.0163moles$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$2KClO_3\rightarrow 2KCl+3O_2$

From the balanced reaction we conclude that,

As, 2 moles of $KClO_3$ react to give 3 moles of $O_2$

So, 0.0163 moles of $KClO_3$ react to give $\frac{3}{2}\times 0.0163=0.0244$ moles of $O_2$

Now we have to calculate the pressure of gas.

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = ?

V = volume of gas = 0.800 L

T = temperature of gas = $25.0^oC=273+25.0=298K$

R = gas constant = 0.0821 L.atm/mole.K

n = number of moles of gas = 0.0244 mole

Now put all the given values in the ideal gas equation, we get:

$P\times (0.800L)=0.0244mole\times (0.0821L.atm/mole.K)\times (298K)$

$P=0.746atm$

Therefore, the final chamber pressure is 0.746 atm.

3 0

4 years ago

How many grams of H3PO4 are in 300 mL of a .50 M solution of H3PO4

Liono4ka [1.6K]

The first thing you need to do is convert mL into L
so (175 mL)(.001L/1mL)=.175L then you multiply by the Molarity of H3PO4 which is 3.5mol/L so (.175L)(3.5mol/L)=.6125 mol H3PO4, and since it wants the answer in grams you then multiply (.6125molH3PO4) by the molar mass of H3P04 which is about 97.99g and your answer is 60.02g which is about 60 grams of H3PO4. Hope this helped

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4 years ago

A gas has a volume of 62.65L at O degrees Celsius and 1 atm. At what temperature in Celsius would the volume of the gas be 78.31

Anastaziya [24]

Answer:

The volume of the gas will be 78.31 L at 1.7 °C.

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