A C5H12 + b O2 → c CO2 + d H2O Find the smallest positive integer values of the stoichiometric coefficients a, b, c and d by setting up a set of linear system of equations from atomic balances and solving the linear system using Gauss elimination and back substitution. Show the details of your work.

Answer

a = 1 ; b = 8 ; c = 5 and d = 6

Explanation

The reactant undergoing combustion reaction is pentane.

First it is important to know the molecular formula for pentane and the products of the combustion reaction so that we can write the unbalanced equation.

C5H12 + O2 → CO2 + H2O

Now we need to balance this equation by ensuring the same number of each atom is on both sides. Starting with carbon, there are 5 on the left hand side but only one on the right hand side so we need to add a 5 in front of the CO2.

C5H12 + O2 → 5CO2 + H2O

Next we need to balance the hydrogens. As there are 12 on the left but only 2 on the right we need to add a 6 in front of the H2O to produce 12 in total (2 x 6 = 12).

C5H12 + O2 → 5CO2 + 6H2O

Finally we need to balance the oxygens. There are 2 on the left but 16 on the right so we need to add an 8 in front of the O2 on the left to produce 16 in total (2 x 8 = 16).

C5H12 + 8O2 → 5CO2 + 6H2O

Calculation approach can also be used.

63.05% of MgCO3.3H2O by mass

Explanation:

of MgCO3.3H2O in the mixture?

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

Mass water:

3.883g - 2.927g = 0.956g water

Moles water -18.01g/mol-

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

Moles MgCO3.3H2O:

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

Mass percent:

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =