Gaseous butane CH3(CH2)2CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 1.31g of water is produced from the reaction of 4.65g of butane and 10.8g of oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

percentage yield of water=

Explanation:

First write the balanced chemical reaction:

mass of buane=4.65g

mass of oxygen=10.8g

mass of water produced=1.31g

mole of butane=

mole of oxygen=

2 moles of butane reacts with 13 moles of oxygen;

so 1 mole of butane reacts with 6.5 moles of oxygen;

0.08 mole will react 0.52 mole of oxygen but we have only 0.34 mole;

therefore butane is excess reagent and oxygen will be the limiting reagent.

amount of product depend on the amount of limiting reagent.

so  13 mole oxygen produce 10 mole of water

hence 1 mole will produce 10/13 mole of water

0.34 mole will give  mole of water.

theoretical mass of water =

but actual produced mass of water =1.31 g

percentage yield of water=

its going to be mutualism cuz they both benefit from the relationship.

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