Chemistry tasks

Solution of chemistry tasks

Solution of chemistry tasks

In practice, we sometimes encounter problems associated with determining a quantitative assessment of the substances entering into a chemical reaction, the estimate of the number obtained in this reaction products, substances of interest to us. The hosts certainly often determine the percentage of a substance in solution and how much one or the other substances you have to take. How to calculate the amount of a substance that would be neutralized or neutralized another substance? How much gas will be allocated, if to the citric acid solution add soda? How to cook, for example, a 5% solution of potassium permanganate, and many other tasks encountered in everyday practice.

The purpose of it article Solution of chemistry tasks - to show how you can easily solve these tasks without resorting to complex chemical calculations, and applying only a general knowledge of mathematics and a bit of general knowledge of chemistry.

What you need to know to solve chemistry tasks

By the way, what You need to know to solve simple chemistry tasks:
M - molar mass (molecular weight substances) - this value for simple substances taken from the periodic table (the number written in the bottom right-hand corner of each element, for example, carbon M(C)=12,01115 g/mol, while the fractional part is usually cast). If it is a gas (e.g. hydrogen), then M(H2) =1 x 2 =2 g/mol), and so for all gases elements.
Basically we are dealing with complex substances, the molar mass of which is equal to the sum of the molar masses of its constituent simple elements, for example, carbon dioxide (CO2): M(CO2) = 12+16x2 = 44 g/mol.
W - concentration is how much of a substance by mass are contained in 100 g of the solution, such as 5% solution contains 5 g of dry matter and 95 g of solvent.

Well, let's look at the most widespread task, how to determine a percentage of a substance in solution and how much you need to take this substance and the solution.

Chemistry task 1

There is 200 ml of 25% acetic acid solution (CH3 -COOH). How much water do you need to take to make a solution of 5% acetic acid from this solution?


Solution) = (М dissolved substance) + (М solvent's)

1) With formula W1 = (M substance) / (М Solution) determine the mass of the dissolved substance:
(M substance) = W1 x (М рsolution), so 0,25 x 200 = 50 g.

2). To get a solution of a lower concentration, you need to dilute it with water, while the mass of the substance dissolved in it will not change. So let's write the same formula for the new solution:
W2 = (M substance) / ( М dissolved substance + М solvent's ). So:
0,05 = 50 / (М solvent's + 50), then М solvent's = 950 g.

3) And so, the mass of the new solution will be equal to (М dissolved substance) + (М solvent's) = 50 + 950 = 1000 г (1л). Knowing the mass of the available solution (200 g) and the mass of the new solution (1000 g) determining that water should be added 800 г.

Sometimes you need to solve the reverse task:

There is a solution of 250 g with a concentration of 5%. It is necessary to determine how much of a soluble substance should be taken to increase the concentration of the solution to 25%?

1) Let's use the formula W1 = (M substance) / (М solution) and determine the mass of the soluble substance in the available 5% solution and the mass of water in the solution:
(M substance) = 0,05 x 250 = 12,5 g., mass (М solvent's) = (М solution) - (M substance) = 250 - 12,5 = 237,5 g.

2) Write down the formula for the new solution:
W2 = (M new substance) / ( М new substance + М solvent's ). Get:
0,25 = (M new substance) / ( М нnew substance + 237,5 ), and М new substance = 79,16 g.

3) So, the previously available 5% solution contained 12.5 g of the dissolved substance, and the new solution contains 79.16 g, so to get a new solution, you need to add 79.16-12.5 g = 66.66 g of the substance.

We decide, count and check!

Mass / volume of the initial solution, g (ml)

Specify % of the available solution

Specify the % of the resulting solution

Chemistry task 2

Task for determining the number of reaction products or reagents.
Let's assume this condition. We need to neutralize the acidic environment. You accidentally spilled 200 ml of hydrochloric acid (HCl), and your have washing soda (Na2CO3) (or its second name is soda ash). So: how much washing soda do you need to take to neutralize 200 ml of hydrochloric acid!?


Write down the equation of the chemical reaction of hydrochloric acid with baking soda:

2HCl + Na2CO3 => 2NaCl + H2O + CO2.

From the course of chemistry, we know that when an acid reacts with a salt, another acid (weaker) and another salt are formed. In our case, carbonic acid is formed, which immediately breaks down into water and carbon dioxide, and a solution of table salt. Now you need to use the periodic table to determine the molecular weight of hydrochloric acid (2 x HCl) and washing soda (Na2CO3). The molecular weight of a complex substance is considered as the sum of the masses of its constituent substances taking into account the number of atoms in the molecule and the number of molecules!
For example, the molecular weight of 2-x molecules HCl: Mr (HCl) = 2 x (1+ 35,5)= 73 g/mol;
molecular weight 1 molecule Na2CO3: Mr (Na2CO3) = 2 х 23 + 12 + 16 х 3)= 106 g/mol
Now it remains to make the proportion:
73 ml (HCl) react with 106 g Na2CO3
200 ml (HCl) react with X g Na2CO3, then X = 200 х 106 / 73 = 290 g.
So, to neutralize 73 ml of spilled acid (concentrated), you need to mix it with 290 g of washing soda. All!

This chemical task may become a little more complicated if you take into account the concentration of acid. In this case, we must calculate how much pure substance is in the solution (as in the previous task 1), and then make a proportion with the amount of hydrochloric acid found. But in any case, even if the acid has a lower concentration, then our solution will satisfy the solution in excess, because, in this case, the soda will be taken in excess!

Chemistry task 3

You need to calculate how much % of each simple substance is in a complex substance - CaSO4 - гипс.


Take the periodic table and find the total relative molecular weight CaSO4: (remember that the total molecular weight of a complex substance is equal to the sum of the molecular weight of its simple substances-choose from the table)
M (CaSO4) = 40 + 32 +16x4 = 136 g/mol
Now we calculate the percentage of each element separately:
40:136 =0,29 (29%)
32:136 = 0,24 (24%)
The percentage of oxygen is found as the remaining part of the task: (100%-29%-24% = 47%).

Chemistry task 4

Have you ever read the expiration date on food packages?! For sure! There is a date of manufacture and shelf life at a certain temperature. But it is not always possible to store food at the recommended temperature. So that's it! You can determine the shelf life of a product by knowing the recommended storage conditions. So, chemistry task:
There is a cake with cream, which has a shelf life of 4 days at a temperature of +50C. Question: how long can this cake be stored at a temperature of +250C?


First of all, when solving such problems (the task of chemistry on the rate of chemical reaction), there is a rule:
when the temperature increases for every 10 0C, the chemical reaction rate increases by 2...4 times
then use formula:

Vt2 = Vt2*Y (t2-t1) / 10,
Vt1 - the rate of the reaction (recommended)
Vt2 - the reaction rate with the new conditions

Y - a coefficient equal to 2, 3, or 4 (in tasks, it is usually 3)
t1 - storage temperature (recommended, indicated on the package)
t2 - storage temperature with a new conditions

So, in our case: t1 = +50C
; t2 = +250C; coefficient Y = 3
then Vt2 / Vt1 = Y(25 - 5) / 10C = 31,5C = 32 = 9.
So, if you store the cake at a temperature of +250C, instead of +50C, then it can be stored in time is 9 times smaller, namely: 4 days = 96 hours, then 96 / 9 = 1.5 hours!