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Makovka662 [10]
4 years ago
14

How do you solve stoichiometry problems

Chemistry
1 answer:
nataly862011 [7]4 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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Explanation:

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See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

M=\frac{n_{solute}}{V_{solution}}

So you can solve for the moles of the solute:

n_{solute}=M*V_{solution}

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

n_{solute}=1mol/L*1.5L=1.5mol

But this is just a supposition.

Regards.

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