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3 years ago

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

1 answer:

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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Solve for Va MaVa=MbVb

PolarNik [594]

Answer:

Va = (MbVb)/Ma

Explanation:

Divide both sides by Ma and voila!

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3 years ago

If an atom has 8 protons. How many electrons does it have? What is the atoms charge?

Juli2301 [7.4K]

The amount if protons is the same as the amount of electrons, so it would be 8. The charge is -2. Look at oxygen in the periodic table.

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3 years ago

Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

Initial: 0.025 0.025

At eqllm: 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago

How do you do this? very confused

Vlad [161]

The problem you have written you almost have it solved. Take the moles that you have calculated and multiply that by the molecular weight to get the grams.

The STP problem:
use the moles you calculated along with 1 atm for Pressure, and 273 for the temperature and plug into the PV = nRT equation. (also use 0.0821 for R)

From there you can solve for the volume

Hope this helps!

4 0

3 years ago

Please I don’t get this at all someone help

ohaa [14]

I think it’s 7.41 because you count up all the atoms and find out how many are x (the large grey ones) and you do 2/27 x 100 which gives you 7.41 :) (sorry if i counted wrong it’s kinda hard)

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3 years ago