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guajiro [1.7K]
3 years ago
12

What is the equation for the reaction between excess dry oxygen and iron fillings​

Chemistry
1 answer:
gtnhenbr [62]3 years ago
6 0

Answer:

Fe2 + O2 > FeO

Explanation:

I don't know how to make the letters small...

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H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles
abruzzese [7]

<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

H_2(g)+F_2(g)\rightarrow 2HF(g)

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]

We are given:

\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = \frac{-14.1}{1}\times 2.20=-31.02J/K

Hence, the value of \Delta S^o for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

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1. How many molecules of ammonia can be created when four molecules of nitrogen are combined with four molecules of hydrogen? In
allochka39001 [22]

Explanation:

1. N_2+3H_2\rightarrow 2NH_3

According to reaction , 1 molecule of nitrogen gas reacts with 3 molecules of hydrogen gas to give 2 molecules of ammonia

Then 4 molecules of nitrogen gas will react with:

\frac{3}{1}\times 4 molecules=12 molecules of hydrogen gas.

As we can see that molecules of nitrogen gas are in excess amount.

So, the molecules  of ammonia  formed will depend upon molecules of hydrogen gas.

According to reaction, 3 molecules of hydrogen gas gives 2 molecules of ammonia.

Then 4 molecules of hydrogen gas will give :

= \frac{2}{3}\times 4 molecules=2.667 molecules of ammonia

2) N_2+3H_2\rightarrow 2NH_3

Molecules of nitrogen gas in balanced chemical equation = 1

Molecules of hydrogen gas in balanced chemical equation = 3

The ratio of nitrogen and hydrogen molecules would result in no left-over reactants will be:

\frac{1}{3} = 1 : 3

For every 1 molecule of nitrogen gas molecules 3 molecules of hydrogen gas molecules are required to form.

3) N_2+3H_2\rightarrow 2NH_3

Moles of nitrogen gas = \frac{100.0 g}{28 g/mol}=3.5714 mol

Moles of hydrogen gas = \frac{100.0 g}{2 g/mol}=50.0 mol

According to reaction , 1 mole of nitrogen gas reacts with 3 mole of hydrogen gas to give 2 moles of ammonia

Then 3.5714 moles of nitrogen gas will react with:

\frac{3}{1}\times 3.5714 mol=10.7142 mol of hydrogen gas.

As we can see that moles of nitrogen gas are only reacting with 10.7142 moles of hydrogen gas which means that nitrogen gas is limiting reagent and hydrogen gas is excessive reagent.

So, amount of ammonia gas will depend upon the moles of nitrogen gas.

According to reaction, 1 molecules of nitrogen gas gives 2 molecules of ammonia.

Then 3.5714 mol of nitrogen gas will give :

= \frac{2}{1}\times 3.5714 mol=7.1428 mol of ammonia

Theoretical yield = Mass of 7.1428 moles of ammonia:

7.1428 mol × 17 g/mol = 199.9 g

199.9 g is the theoretical yield of the reaction.

Hydrogen gas  is the excess reactant.

Nitrogen gas is the limiting reactant.

7 0
3 years ago
Read 2 more answers
The forensic technician at a crime scene has just prepared a luminol stock solution by adding 17.0 \({\rm g}\) of luminol into a
maria [59]

Answer:

1.28 M

Explanation:

Step 1: Given data

Mass of luminol (solute): 17.0 g

Volume of water: 75.0 mL (this is also the volume of solution)

Step 2: Calculate the moles corresponding to 17.0 g of luminol

The molar mass of luminol is 177.16 g/mol.

17.0 g × 1 mol/177.16 g = 0.0960 mol

Step 3: Calculate the molarity of the solution

We will use the definition of molarity

M = moles of solute / liters of solution

M = 0.0960 mol / 0.0750 L = 1.28 M

3 0
3 years ago
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