Nasri XII Multimedia

Chemistry Counts


Count the ways chemistry is calculation oriented fundamental laws of chemistry.

In this case will be given an assortment of sample sums and their chemistry pembahasanya.

Examples of questions:
1.What percentage of the levels of calcium (Ca) in calcium carbonate? (Ar: C = 12; O = 16; Ca = 40)
Answer:
1 mol CaCO, containing 1 mol Ca + 1 + 3 mol C mol O
Mr CaCO3 = 40 + 12 + 48 = 100
So the level of calcium in CaCO3 = 40/100 x 100% = 40%


2.A total of 5.4 grams of aluminum metal (Ar = 27) was treated with excess dilute hydrochloric acid according to the reaction:

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

How many grams of aluminum chloride and how many liters of hydrogen gas produced at standard conditions?
Answer:
Of the equation can be expressed
2 mol Al x 2 mol AlCl3 ® 3 mol H2
5.4 grams of Al = 05/04/27 = 0.2 mol
So:
AlCl3 formed = 0.2 x Mr AlCl3 = 0.2 x 133.5 = 26.7 grams
The volume of H2 gas produced (0o C, 1 atm) = 3/2 x 0.2 x 4.22 = 6.72 liters


3.An iron ore contains 80% Fe2O3 (Ar: Fe = 56, O = 16). Oxide is reduced with CO gas to produce iron.
How many tons of iron ore required to make 224 tons of iron?

Answer:
1 mol Fe2O3 contains 2 mol Fe
then: mass Fe2O3 = (Mr Fe2O3 / 2 Ar Fe) Fe = mass x (160/112) x 224 = 320 tons
So the iron ore required = (100/80) x 320 tons = 400 tons


4.To determine the crystal water 24.95 grams of copper sulfate salt crystal is heated until all the water evaporates. After heating the salt mass to be 15.95 grams. How much water is contained in the crystal salt?

Answer:
suppose formula is CuSO4 salt. xH2O

CuSO4. xH2O ® CuSO4 + xH2O

24.95 grams of CuSO4. xH2O + 18x = 159.5 moles

15.95 grams CuSO4 = 159.5 mol = 0.1 mol

according to the above equation can be stated that:
number of moles of CuS04. CuSO4 xH2O = mol; thus equation

24.95 / (159.5 + 18x) = 0.1 ® x = 5

So the formula is CuS04 salt. 5H2O



Empirical Formula and Molecular Formula

The empirical formula is the simplest formula of a compound.
This formula simply states ratio of the number of atoms contained in the molecule.
The empirical formula of a compound can be determined if known one:
- Ar mass and each element
-% Ar mass and each element
- Ar mass ratio and each element

Molecular formula: if the empirical formula is known and it is also known to Mr formulas can be determined.
Example: A compound C den H contains 6 grams of C and 1 gram H.
Determine the empirical formula and the molecular formula of the compound where it is known Mr = 28!
Answer:

mol C: mol H = 6/12: 1/1 = 1/2: 1 = 1: 2
So the empirical formula: (CH2) n

When Mr compound = 28 then: 12n + 2n = 28 ® 14N = 28 ® n = 2

So the molecular formula: (CH 2) 2 = C2H4

Example: For a 20 ml oxidize hydrocarbons (CxHy) required oxygen in a gaseous state as much as 100 ml and 60 ml of CO2 produced. Determine the molecular formula of the hydrocarbon?
Answer:

Hydrocarbon combustion equation in general

CxHy (g) + (x + 1/4 y) O2 (g) ® x CO2 (g) + 1/2 y H2O (l)
The coefficient indicates the reaction mole ratio of the substances involved in the reaction.
According to Gay Lussac gases on p, t the same time, the number of moles is proportional to the volume

Then:
CxHy mol: mol O2: mol CO2 = 1: (x + 1/4y): x
20: 100: 60 = 1: (x + 1/4y): x
1: 5: 3 = 1: (x + 1/4y): x

or:

1: 3 = 1: x ® x = 3
1: 5 = 1: (x + 1/4y) ® y = 8
So hydrocarbon formula is: C3H8