### reaction calculations

Whereever you apply chemistry, for example in a lab or in a factory, making a product needs knowledge of the needed
How much product you want? How much reactant is needed therefore? In what proportion you must mix the components?

So, well calculating before starting is not only important to get the right product, but shows also your economic understanding. It is stupid to use too much of any reactant.

The application of reaction calculations can only be done by someone who knows to deal with the concept of MOL (see also module 5).
It is very important to be capable to make calculation conversion from MOL to grammes and v.v. Besides that, you must know what density is of solids and liquids.

Remeber the following equation: d = g/v?

Gases also have a density, but with a very different definition than density of (s) and (l). In a next paragraph we will meet that.

Already good to know about gases: whatever gas you take, with big or small molecules, one mol of it always has the same volume (measured at the same temperature and pressure). Or the other way round? if you take 1 lieter of a gas, measured at a fixed temperature and pressure, this 1 liter has the same number of moles, no matter what gas you take.
MOL and volume of gases are directy connected. What is impossible with solids or liquids, can be done with gases: the coefficients of gases in a reaction equation can be read as MOL and as liters.
The mol proportion = de volume proportion.

If methane reacts with Oxygen in the proportion 1:2 (one mol methane reacts with 2 mol Oxygen), this means that simultaneously and automatically, under the same conditions, 1 liter of methane gas reacts with 2 liter Oxygen gas.

You must know the concepts of "molarity" and concentration. You will meet these very often in reaction calculations.

For example: 10 ml 0,1M HCl reacts with one or another substance. You must be able to apply these data in the calculation. You must know very well this way of writing.

A reaction calculation always starts with a well balanced reaction equation, that clearly indicates the mol proportion.

The calculation in question 13 is an example to show the advantage of using empiric formulas instead of ionic formulas.
Calculations deal with real and practical things and in practice you don't get ions from a bottle, but complete neutral substances.
So, even if only Chloride ions participate in the reaction, you will measure salt (NaCl) on your balance.

Again: it is wise, for good understanding of the practice, to indicate if the substances are (s), (l) or (g) (or (aq)).
Ca(s) + 2H2O (l) Ca(OH)2(aq) + H2(g)
The reactants react in the mol proportion 1 : 2

Att. That mol proportion shows in what proportion the substances react; that can be very different than the composition wherein the substances were mixed in reality. If you do not mix as well as possible in the right mol proportion, unused substance will remain (and complicate).

 Important rules for resolving a reaction calculation: 1 Use balanced reaction equation and add the phases. 2 Underline the data, the given data and those asked for. The other substances (without any data) are not needed in the calculation. Sometimes the data are directly given, sometimes indirectly. The calculation is continues with only the underlined substances. 3 write down the mol proportion 4 Where needed, convert the mol into the right units (to be found in the data and in the required answer) 5 Introduce a conversion factor to arrive at the real amounts, given in the data. That's how you finish the calculation.
Att.: It is absolutely necessary that you continues in every step with all (underlined) substances, to assure that you know what you are doing. Only then the calculation remains a real thing, and not an abstractum.

EXAMPLE:
 Calculate mass and volume (standaerd conditions for gases) of carbon dioxyde produced in the complete combustion of 4,01 g methane? 1 CH4(g) + 2 O2(g) CO2(g) + 2H2O(g) 2 Underline the substances with data. CH4(g) + 2 O2(g) CO2(g) + 2H2O(g) 3 So, 1 mol CH4(g) reacts with 1 mol CO2(g) (proportion 1:1) 4 16 gramme CH4(g) produce 44 gramme CO2(g) (here we apply the molecular masses) 5 in reality we do not have 16 grammes, but only 4 grammes for combustion. To be introduced a factor of 4/16. (in this case we can simply divide by 4) 4/16 x 16 gramme CH4(g) produce 4/16 x 44 gramme CO2(g) to finish: standard condition mean: at temp = 25oC and pressure of 1 atm. Then 1 mol gas = 22,4 liter 1/16 x 44 = 11 gram CO2(g) is produced, that equals 4/16 mol = 4/16 x 22,4 liter CO2(g) = 5,6 liter

Again the rules for resolving the reaction equation:
1. A balanced reaction equation + phases.
2. Underline the substances about which you have data or you must find answers.
3. Note the MOL proportion
4. If needed, convert MOL into other units.
5. Introduce a calculation factor, to reach the real amounts (see data).