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3 years ago

What is the molar ratio for the following equation after it has been properly balanced?

Chemistry

1 answer:

hichkok12 [17]3 years ago

6 0

6 Na + 1 Fe₂O₃ → 3 Na₂O + 6 Fe

<h3>Explanation</h3>

Method One: Refer to electron transfers.

Oxidation states:

Na: from 0 to +1; loses one electron.
Fe: from +3 to 0; gains three electrons.

Each mole of Fe₂O₃ contains two Fe atoms and will gain 2 × 3 = 6 electrons during the reaction. It takes 6 moles of Na to supply all those electrons.

6 Na + 1 Fe₂O₃ → ? Na₂O + ? Fe

There are two moles of Na atoms in each mole of Na₂O. 6 moles of Na will make 3 moles of Na₂O.
There are two moles of Fe atoms in each mole of Fe₂O₃. 1 mole of Fe₂O₃ will make 2 moles of Fe.

6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe

Method Two: Atoms conserve.

Fe₂O₃ has the largest number of atoms among one mole of all four species in this reaction. Assume one as its coefficient.

? Na + 1 Fe₂O₃ → ? Na₂O + ? Fe

There are two moles of Fe atoms and three moles of O atoms in each mol of Fe₂O₃. One mole of Fe₂O₃ contains two moles of Fe and three moles of O. There are one mole of O atom in every mole of Na₂O. Three moles of O will go to three moles of Na₂O.

? Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe

Each mole of Na₂O contains two moles of Na. Three moles of Na₂O will contain six moles of Na.

6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe

Simplify the coefficients. All coefficients in this equation are now full number and relatively prime. Hence the equation is balanced.

6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe

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If the lab technician needs 30 liters of a 25% acid solution, how many liters of the 10% and the 30% acid solutions should she m

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Answer:

7.5 L of the 10% and 22.5 L of the 30% acid solution, she should mix.

Explanation:

Let the volume of 10% acid solution used to make the mixture = x L

So, the volume of 30% acid solution used to make the mixture = y L

Total volume of the mixture = x + y = 30 L .................. (1)

For 10% acid solution:

C₁ = 10% , V₁ = x L

For 30% acid solution :

C₂ = 30% , V₂ = y L

For the resultant solution of sulfuric acid:

C₃ = 25% , V₃ = 30 L

Using

C₁V₁ + C₂V₂ = C₃V₃

10×x + 30×y = 25×30

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x + 3y = 75 .................. (2)

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3 years ago

Read 2 more answers

The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha

zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$25.9 = x*10.0$

$25.9 = 10.0\:x$

$10.0\:x = 25.9$

$x = \dfrac{25.9}{10.0}$

$\boxed{x = 2.59}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{53.3}{2^{2.59}}$

$m \approx \dfrac{53.3}{6.021}$

$\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

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3 years ago

above shows a balloon full of gas which has a volume of 120.0ml at 300.0k assuming pressure remains constant , what is the voume

trasher [3.6K]

128 ml is the voume of the balloon if the temperature of the gas increases to 320.0k.

Explanation:

given that:

T1 (initial temperature) = 300K

V1 ( initial volume) = 120ml

T2 (final temperature) = 320 K

V2 (final volume) = ?

Pressure remained constant throughout the process.

From the equation

$\frac{P1V1}{T1}$ = $\frac{P2V2}{T2}$

Since pressure is constant the equation will be:

$\frac{V1}{T1}$ = $\frac{V2}{T2}$

V2 = $\frac{V1 X T2}{T1}$

Putting the values in the above formula:

V2 = $\frac{120 X 320}{300}$

= 128 ml

128 ml is the volume of the gas if temperature increases from 3OO K to 320k

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