Question

Find the mass of ${\rm{C}}{{\rm{O}}_{\rm{2}}}$ produced when 5.8 g of butane ${{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}$ is burnt. (C=12, O=16, H=1)

$2{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}} + 13{{\rm{O}}_{\rm{2}}} \to 8{\rm{C}}{{\rm{O}}_{\rm{2}}} + 10{{\rm{H}}_{\rm{2}}}{\rm{O}}$

Answer 1

Hint: We know that a chemical reaction when written in the balanced form gives a quantitative relationship in terms of molecules, masses, moles and volume between the different reactants and products involved in the reactions. This is termed as stoichiometry.

Complete step by step answer:
Here, Butane reacts with oxygen to form carbon dioxide and water. The chemical reaction is given as,

$2{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}} + 13{{\rm{O}}_{\rm{2}}} \to 8{\rm{C}}{{\rm{O}}_{\rm{2}}} + 10{{\rm{H}}_{\rm{2}}}{\rm{O}}$

Given that, the mass of butane burnt is 5.8 g. The molar mass of butane is given as 58 ${\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$, molar mass of oxygen is 32 ${\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$. Molar mass of ${\rm{C}}{{\rm{O}}_{\rm{2}}} = 12 + 2 \times 16 = 44\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$. Now, we have to use stoichiometry to calculate the mass of carbon dioxide forms.

Using stoichiometry,

${\rm{Mass}}\,{\rm{of}}\,{\rm{C}}{{\rm{O}}_{\rm{2}}} = 5.8\,{\rm{g}}\,{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}} \times \dfrac{{1\,{\rm{mol}}\,{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}}}{{{\rm{58}}\,{\rm{g}}\,{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}}} \times \dfrac{{8\,{\rm{mol}}\,{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{2\,{\rm{mol}}\,{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}}} \times \dfrac{{44\,{\rm{g}}\,{\rm{C}}{{\rm{O}}_{\rm{2}}}}}{{1\,{\rm{mol}}\,{\rm{C}}{{\rm{O}}_2}}}$

$ \Rightarrow {\rm{Mass}}\,{\rm{of}}\,{\rm{C}}{{\rm{O}}_{\rm{2}}} = \dfrac{{5.8 \times 8 \times 44}}{{116}} = 17.6\,{\rm{g}}$

Therefore, burning of 5.8 of butane produces 17.6 g of carbon dioxide.

Additional Information:
Stoichiometry calculations are of three types:

1) Involving mass-mass relationship: In this type of calculations, mass of one reactant or product is given and that of the other can be calculated.

2) Involving mass-volume relationship: In this type of problem, the mass/ volume of one of the reactants or products is given and that of the other can be calculated.

3) Involving volume –volume relationship: In this case, the volume of one of the reactants or products is given and that of the other can be calculated.

Note: Always remember to balance the chemical reaction before proceeding to stoichiometry calculation. To calculate stoichiometry by mass, first we have to express the reactant in terms of moles and then should be multiplied with the molar mass of each to give the masses of each reactant per mole of reaction.