reaction calculations

When, long time ago, Jules Verne wrote his great voyage to the moon, he described exactly how much gun powder was needed to empower the "rocket" with the crew to leave the earth, how much of the chemicals should be taken with them, the exact amount of Oxygen needed for those men during the planned days of the trip, etcetera.
If today something like that really happens, a similar calculation has still to be done with the same accuracy or better.
For example: a solid rocket fule is hydrazine, made of ammonia (to be calculated in liters of gas) and Sodium hypochlorite (to be calculated in grammes. Toghether they react and produce the right amount of hydrazine and no reactants are left over.

Another example:
If in a sulphuric acid works, sulphur dioxyde(g), oxygen(g) and water(l) is mixed to produce sulphuric acid, then the amount of every needed reactant is calculated very precisely.

There are so many applications. Any factory starts the production with a well balanced reaction equation in order to calculate how much product can be made from how much reactant.
Not doing so, mixing some reactants arbitrarily, gives the big chance that wrong products and side products wil appear, and secondly, that much reactant gets lost; economically (and environmentally) this would be desastrous, cost a lot of money, etc.
The good manager looks for someone that knows about reaction calculations.


General


Demonstration at school, or in a lab:
x g Magnesium is dissolved in 1M HCl.
The produced gas is collected and measured in a gas seringe.
Starting with that known amount of gas, measured at room temperature, must be calculated how much Magnesium was dissolved in the acid.

Preparing a chemical reaction, it is important the you know how much of the reactants is needed and how much of the products you can expect. This is not only a chemical, but also an economic interest.

Always atoms, molecules, ions collide when reacting; these particles can be very different in size and in mass (but not necessarily).
Thus a big and heavy atom can very well react with a very tiny ion. Of course this is reflected in the 'mass proportion'.

For every reaction one thing is clear: whatever amount of atoms, molecules or ions you bring toghether, they will always react in a fixed proportion. That fixed number can be read in the coefficients of the balanced reaction equation.
The coefficients in the reaction equation have two significations:
    In the reaction: 2H2 + O2 2H2O the coefficients mean to say:
  1. (micro): 2 molecules of Hydrogen react with 1 molecule of Oxygen and produce 2 molecules of water.
  2. (macro): 2 mol of Hydrogen react with 1 mmol of Oxygen and produce 2 mols of water.
The proportion 2 : 1 : 2 is a fixed one and is called: the mol proportion.
A reaction calculation always starts with a well balanced reaction equation.

Note that the proper reaction determines the coefficients. The coefficiebnts do not depend on the amount of substances available.
Even if you mix 100 mol Hydrogen with 1 mol of Nitrogen, the mol proportion of this reaction wil remain the same:

N2(g) + 3H2(g) 2NH3(g)
so
1 : 3 : 2

Substances built up of ions (mostly to be recognized with the fact that there is a metal present in the compound) in dissolved state, are split up, dissociated into ions; in solid state they have an ionic lattice.
In reaction calculations mostly counts the amount of not dissociated substance, needed or produced. That's why equations for calculations are written in proportion formulas, not in ions. that is easier for the calculation.
You do not weigh loose ions, don't you. We don't have bottles with only Sodium ions.


We write in ions: Na+ + Cl- + Ag+ + NO3- Na+ + NO3- + AgCl(s)

We write in proportion formula: NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl(s)

In chemical calculations, the last way of writing is preferred.
Always add the state of matter (the phase): (s) (l) (g) (aq). This can prevent many mistakes.

The reactants react in a fixed mol proportion; when they are not mixed in the right proportion, they react as long as there still are both reactants, until one of them is totally consumed.
The rest remains unchanged.

For gases and for liquids we often use a unit of volume: liters or mls.
Transforming these units into other units might be necessary. We often use the 'density' to transform volume into mass or v.v. The volume of a gas can also directly be translated into mol (knowing the t and p).

It is wise, for good understanding of the practice, to indicate the state/phase (s), (l) of (g) (or (aq)):

Ca(s) + 2H2O(l) Ca(OH)2(aq) + H2(g)

This calculation is also an example to show the advantage of using the proportion formula (or empirical formula) more than ionic formulas.
Chemical calculations deal with real stuff, and ions alone cannot be dealt with.
So even when only the Chloride ions participate, yet you must weigh kitchen salt (NaCl).

N.B. Remember the equation: d = g/v?

Gases also have a density, but gas density has a complete different definition than density of (s) or (l).
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 t and p.
In other words:
If you take on liter of a gas, measured at a fixed t and p, than whatever gas you take, that liter contains the same number of mols.
Mol and volume are proportional in the case of gases.
What is absolutely not true for liquids and solids, is true for gases: the coefficients of gases in an equation can be read as mols ánd as liters or ml.

For gases: MOL PROPORTION = VOLUME PROPORTION

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



Example:
What is the mass and the volume (standard circumstances) of Carbon dioxyde produced in the complete combustion of 4.01 g Methane?
1 CH4(g) + 2 O2(g) CO2(g) + 2H2O(g)
2 Subline those substances about which you have data, or about which is a question. CH4(g) + 2 O2(g) CO2(g) + 2H2O(g)
3 So, 1 mol CH4(g) reacts with 1 mol CO2(g) (proportion is 1:1)
4 16 g CH4(g) produce 44 g CO2(g) (here we apply the molecular masses)
5 in reality we have not 16 g, but only 4 g to be burnt.
The factor to be introduces is: 4/16.
4/16 x 16 g CH4(g) produce 4/16 x 44 g CO2(g)

finally: standard circumstances mean: at temp 25°C and pressure 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