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sattari [20]
3 years ago
9

5.4 g of Aluminium reacts with 300 mL of 0.2 mol/L hydrochloric acid solution. A. Write equation for the reaction taking place.

B. Specify which reactant is limiting and which reactant is excess. C. Find volume of the gas collected at S.T.P d. How many grams of salt are produced at the end of the reaction? E. How many grams of the excess reactant are left ate the end of the reaction? Given: Al=27 , H=1 , Cl=35.5 Chemistry grade 10
Chemistry
1 answer:
insens350 [35]3 years ago
5 0

Answer:

A. 2 Al (s) + 6HCl (aq) →  2AlCl₃ (s) ↓ + 3H₂ (g)  

B. Al is the excess reactant and HCl is the limiting.

C. 0.672 L of H₂ produced at STP

D. 2.67 g of AlCl₃ are made in this reaction.

E. 4.86 g of Al remain after the reaction goes complete.

Explanation:

We star from the reaction:

2 Al (s) + 6HCl (aq) →  2AlCl₃ (s) ↓ + 3H₂ (g)  

2 moles of aluminum, react with 6 moles of HCl in order to produce 2 moles of aluminum chloride and 3 mol of H₂ gas.

We determine moles of each reactant:

[HCl] = 0.2M → 0.2 mol/L . 0.3L = 0.060 moles

(we converted 300 mL to 0.3L)

5.4 g of Al . 1mol / 26.98g = 0.200 moles

Ratio is 2:6 (3). 2 mol of Al react to 6 mol of HCl

0.2 moles of Al may react with (0.2 . 6) /2 = 0.6 mol of acid

We have 0.06 moles, and we need 0.6 mol of acid, so the HCl is the limiting reactant. Then, the Al is the excess:

6 moles of HCl need 2 moles of Al to react

Then 0.06 moles of HCl will react to (0.06 . 2) /6 = 0.02 moles

If we have 0.2 moles of Al, and we need 0.02 moles for the reaction, then

(0.2 - 0.02) = 0.18 moles remain after the reaction is complete.

0.18 mol . 26.98g /1mol = 4.86 g of Al remain after the reaction goes complete.

As the limting reactant is the HCl, we work with it to determine the mass of salt which is produced:

6 mol of HCl can produce 2 mol of chloride

Then 0.06 moles of HCl will produce (0.06 . 2) /6 = 0.02 mol of AlCl₃

We convert to mass: 0.02 mol . 133.33g/1mol = 2.67 g of AlCl₃ are made in this reaction.

Let's find out the volume of hydrogen produced, at STP

6 moles of HCl can produce 3 moles of H₂

0.06 moles of HCl will produce (0.06 . 3) /6 = 0.03 moles of H₂

1 mol of any gas at STP occupies 22.4L

0.03 moles of H₂ will ocuppy (22.4 L . 0.03 mol)/1mol = 0.672L

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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:

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We have 2 grams of H2
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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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