How to solve chemistry problems 8th grade

In most cases, current or former students there are at least some theoretical understanding of chemical processes. But here is the solution of tasks on chemistry is quite a difficult situation if there are certain skills. But chemical task is to help in the kitchen when breeding, for example, acetic essence or just a friendly hint own son or sister. Remember how to solve chemistry problems? Usually, in the 8th grade, the first problem using the equations of chemical reactions are of the type "Calculate the mass of one of the reaction products by a known mass of one of the reacting substances. The problem is solved with the help of chemical formulas, because often jobs use this method.

How to solve chemistry problems 8th grade
How to learn to solve chemistry problems
How to solve ionic equations
How to find the equivalent mass of

Instruction

Task. Calculate the mass of aluminium sulfide if the reaction of sulfuric acid entered the 2.7 g of aluminium.

Recorded a short condition

This:

m(Al) =2, 7 g

H2SO4

Find:

m(Al2 (SO4) 3)-?

Before to solve problems in chemistry, a chemical equation. In the interaction of metals with dilute acid forming salt and released gaseous substance is hydrogen. We arrange the coefficients.

2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2

When you always need to pay attention only to substances for which known and you need to find, options. All the rest are not taken into account. In this case it will be: Al and Al2 (SO4) 3

Find the relative molecular mass of these substances in the table of D. I. Mendeleev

Mr(Al) =27

Mr(Al2 (SO4) 3) =27•2(32•3+16•4•3) =342

Translate these values into molar mass (M) multiplied by 1 g/mol

M(Al) =27G/mol

M(Al2 (SO4) 3) =342g/mol

Recorded a basic formula, which connects the amount of substance (n), mass (m) and molar mass (M).

n=m/M

Carry out calculations according to the formula

n(Al) =2.7 g/27G/mol=0.1 mol

Make up two ratio. The first relation based on equation based on the odds faced by the formulas of the substances whose parameters are given or need to find.

First ratio: 2 mol Al you have 1 mole of Al2 (SO4) 3

The second ratio of 0.1 mol Al we have X mole Al2 (SO4) 3

(compiled on the basis of the obtained results)

Solve the proportion, given that X is the amount of substance

Al2 (SO4) 3 and has the unit mol

Here

n(Al2 (SO4) 3)=0.1 mol(Al)•1 mol(Al2 (SO4) 3):2моль Al=0.05 mol

Now there is the amount of substance and molar mass of Al2(SO4)3, consequently, you can find lots of which derive from the basic formula

m=nM

m(Al2 (SO4) 3)=0.05 mol•342g/mol=17,1 g

Recorded

Answer: m(Al2 (SO4) 3)=17,1 g

At first glance, it seems that to solve chemistry problems is very difficult, but it is not. And to verify the degree of assimilation, first try to solve the same problem, but only yourself. Then substitute other values using the same equation. And the last step is the solution of the new equation. And if you were able to handle what – you're to be congratulated!

Useful advice

Wonderful helper during solving tasks is a manual, time-tested "problems in chemistry for entering Universities" G. P. Khomchenko. And don't be afraid to use it – it proposes the solution of problems from the very beginning!

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