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QuestionCalculate the mass of water produced when 1.234 g1.234 \mathrm{~g} of butane reacts with 4.416 g4.416 \mathrm{~g} of oxygen, generating 3.737 g3.737 \mathrm{~g} of CO2\mathrm{CO}_{2}.

Studdy Solution

STEP 1

Assumptions1. The mass of butane burned is1.234 g. The mass of oxygen reacted is4.416 g3. The mass of carbon dioxide formed is3.737 g4. The reaction is complete and all of the butane and oxygen are used up5. The only products of the reaction are carbon dioxide and water

STEP 2

We can use the law of conservation of mass to solve this problem. The law of conservation of mass states that the total mass of the reactants is equal to the total mass of the products in a chemical reaction.
Mass of reactants=Mass of productsMass\, of\, reactants = Mass\, of\, products

STEP 3

Now, plug in the given values for the mass of the reactants and the mass of carbon dioxide to find the mass of water.
Mass of reactants=Mass of CO2+Mass of H2Mass\, of\, reactants = Mass\, of\, CO_{2} + Mass\, of\, H_{2}

STEP 4

Substitute the known values into the equation.
1.234 g (butane)+4.416 g (oxygen)=3.737 g (CO2)+Mass of H21.234\, g\, (butane) +4.416\, g\, (oxygen) =3.737\, g\, (CO_{2}) + Mass\, of\, H_{2}

STEP 5

Rearrange the equation to solve for the mass of water.
Mass of H2=1.234 g (butane)+4.416 g (oxygen)−3.737 g (CO2)Mass\, of\, H_{2} =1.234\, g\, (butane) +4.416\, g\, (oxygen) -3.737\, g\, (CO_{2})

STEP 6

Calculate the mass of water.
Mass of H2=1.234 g+4.416 g−3.737 g=1.913 gMass\, of\, H_{2} =1.234\, g +4.416\, g -3.737\, g =1.913\, gThe mass of water formed in the reaction is1.913 g.

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