Question

How many grams of Co, are produced when 88 g of o, is reacted with an excess of
butane?

Answer 1

`74.5gCO_2`

Step-by-step explanation:

Hello there!

In this case, for the described reaction, it is possible to realize there is a 13:8 mole ratio between oxygen (O2) and carbon dioxide (CO2); moreover, since the molar mass of the former is 32.00 g/mol and that of latter is 44.01 g/mol, the produced mass of the required product turns out to be:

`88gO_2*(1molO_2)/(32.00gO_2)*(8molCO_2)/(13molO_2)*(44.01gCO_2)/(1molCO_2) \\\\=74.5gCO_2`

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