Instruction
1
Task. Calculate the mass of aluminium sulfide if the reaction of sulfuric acid entered the 2.7 g of aluminium.
2

Recorded a short condition

This:

m(Al) =2, 7 g

H2SO4

Find:

m(Al2 (SO4) 3)-? 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 4

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 5

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 6

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 7

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 