Question

Calculate the mass of water produced when 7.81 g of butane reacts with excess oxygen. 2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g)

Answer 1

12.09g of water are produced.

Step-by-step explanation:

1st) From the chemical equation we know that 2 moles of C4H10 (butane) react with oxygen to preoduce 10 moles of H20 (water) and CO2.

It is necessary to use the molar mass of C4H10 (58.12g/mol) and H2O (18g/mol) to convert moles to grams:

• C4H10 conversion:

`2moles*(58.12g)/(1mole)=116.24g`

• H2O conversion:

`10moles*(18g)/(1mole)=180g`

Now we know that 180g of water are produced from 116.24g of butane.

2nd) With a mathematical rule of three and the stoichiometry of the reaction we can calculate the mass of water produced when 7.81g of butane reacts with excess oxygen:

`\begin{gathered} 116.24gButane-180gWater \\ 7.81gButane-x=(7.81gButane*180gWater)/(116.24gButane) \\ x=12.09gWater \\ \end{gathered}`

Finally, 12.09g of water are produced.