Gaseous butane ch3ch22ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o. suppose 24. g of butane is mixed with 44.3 g of oxygen. calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. round your answer to 3 significant digits.

Maximum amount of can be produced is 37.5 g

Explanation:

Balanced equation:

Molar mass of butane ()  = 58.12 g/mol

Molar mass of = 32 g/mol

Molar mass of = 44.01 g/mol

So, 24 g of butane  = of butane = 2.422 mol of butane

Also, 44.3 g of  = of = 1.384 mol of

According to balanced equation-

2 moles of butane produce 8 mol of

So, 2.422 moles of butane produce of = 9.688 moles of

13 moles of produce 8 mol of

So, 1.384 moles of produce of = 0.8517 moles of

As least number of moles of are produced from therefore is the limiting reagent.

So, maximum amount of can be produced = 0.8517 moles =

5.6⋅10−^19  j

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