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Makovka662 [10]
3 years ago
14

How do you solve stoichiometry problems

Chemistry
1 answer:
nataly862011 [7]3 years ago
7 0
Ok so the way I do it is as simple as possible.
Firstly look at the reactants and products ( there can be one reactant and one product or more ) you will usually be given the moles of the reactant or products, if you are given grams you can convert into moles by this convertion ( grams/R.M.M ) where R.M.M is the relative atomic mass of your substance ( the mass number of all of the elements in your substance).

Ok when you have moles now look at the ratio between the products and reactants. Usually you will won't know the moles of one substance therefore you will be asked to find moles or mass of that substance.

For example:

When 16 grams of oxygen and 1 gram of hydrogen gas react to produce water. Find the number of grams of water being produced.

O2 + 2H2 -> 2H2O
16g      2g          xg

Here we're told the mass of the reactants. In stoichiometry we need to work with moles therefore you need to calculate moles of the reactants.
Firstly find the R.M.M of each reactant.
R.M.M of O2 is 16+16=32 since it's diatomic we add atomic masses of two oxygen atoms.
R.M.M of H2 is 1+1=2, it's also diatomic. (Diatomic two atoms of the same element are joined together). (Ignore the number 2 in front of H2, this number shows us the ratio relationship between reactans or products, i.e when we balance an equation.)

Ok so now find moles:

We have 16 grams of O2
16/R.M.M
16/32 = 0.5 moles

We have 2 grams of H2
1/R.M.M
2/2 = 1 mole
Now back to the equation.

O2 + 2H2 -> 2H2O
0.5 moles 1mole xmoles                  (it's xmoles because we don't know                                                               molarity of water that's what we have                                                           find firstly in order to find grams.)
Now look at the ratio between any reactant and product i.e you can choose which reactant to compare to the product, it doesn't make a different ( I will do two or you can do two at the same time)

1st method:
Look at the ratio between O2 and H2O from the reaction above we see the ratio is 1:2 therefore for every 0.5 moles of O2 you get 1 mole of H2O.
1:2
0.5 : x
0.5*2 = 1

2nd method;
Look at the ratio betweem H2 and H2O from the reaction above we see the ratio is 2:2 or 1:1. We have 1 mole of H2 there we must have 1 mole of H2O. We see this is true as both methods give us 1 mole of H2O.

3rd method ( combined):
Look at the ratio between O2, H2 and H2O.
We see that the ratio is 1:2:2
So we have 0.5:1:x
If we multiply 0.5 *2 it equals 1 mole
If we multiply 1*1 we get 1 moles.
Any method is correct and it's up to you to find a comfortable way.
We're not finished in the question we are asked for the mass of water.
So just multiply the number of moles (1mole) by R.M.M of H2O.
1 * R.M.M
R.M.M of H2O = 1+1+16=18
1*18= 18 grams.
And you're finished.

I am sorry if this is so long I want you to understand as much as possible.
In stoichiometry you can also be asked about the empirical formula of a substance. I can show you how do it. If you have any question just tell me.
Hope this helps :).
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How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?
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Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

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What scientist described a positively charged core in the middle of the atom
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8 0
3 years ago
Methane, the principal component of natural gas, is used for heating and cooking. The combustion process is CH4(g) + 2 O2(g) → C
antoniya [11.8K]

Answer:

392.97 litres

Explanation:

From the equation of reaction, we can see that 1 mole of methane yielded 1 mole of carbon iv oxide. Hence, 15.9 moles of methane will yield 15.9 moles of carbon iv oxide.

At s.t.p one mole of a gas occupies a volume of 22.4L ,hence 15.9 moles of a gas will occupy a volume of 22.4 × 15.9 which equals

356.16L.

Now, we can use the general gas equation to get the volume produced at the values given.

We have the following values;

V1 = 356.16L P1= 1 atm ( standard pressure) T1 = 273K ( standard temperature) V2 = ? T2 = 23.7 + 273 = 296.7K P2 = 0.985 atm

The general form of the general gas equation is given as :

(P1V1)T1 = (P2V2)/T2

After substituting the values , we get V2 to be 392.97Litres

8 0
2 years ago
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