In a car gas tank, 46.0 g of gasoline burns consuming 173 g of oxygen. The resultant products are carbon dioxide are carbon dixide and water. What is the total combined mass of the products?
Would it be 173g - 46.0g = 127g
46.0 g gasoline + 173 g oxygen = total mass to start of 219.0 grams. Since CO2 and H2O are the only products formed, their total weight must be 219.0 grams. Law of conservation of mass. Check my arithmetic.
Oh! I get it thanks!
To find the total combined mass of the products, we need to consider the law of conservation of mass, which states that mass is neither created nor destroyed during a chemical reaction. In this case, the reactants are gasoline and oxygen, and the products are carbon dioxide and water.
The combustion of gasoline can be represented by the balanced chemical equation:
C₈H₁₈ (gasoline) + 12.5O₂ (oxygen) -> 8CO₂ (carbon dioxide) + 9H₂O (water)
From the equation, we can see that for every 1 mole of gasoline (114.22 g), we need 12.5 moles of oxygen (12.5 * 32.00 g) to react completely, resulting in 8 moles of carbon dioxide (8 * 44.01 g) and 9 moles of water (9 * 18.015 g).
Now, let's calculate the moles of gasoline and oxygen in the given problem.
Moles of gasoline:
Mass of gasoline = 46.0 g
Molar mass of gasoline = 114.22 g/mol
Moles of gasoline = Mass of gasoline / Molar mass of gasoline
Moles of oxygen:
Mass of oxygen = 173 g
Molar mass of oxygen = 32.00 g/mol
Moles of oxygen = Mass of oxygen / Molar mass of oxygen
Next, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
To find the limiting reactant, compare the moles of each reactant to the stoichiometry mentioned in the balanced equation. The reactant with fewer moles is the limiting reactant.
Then, using the stoichiometry from the balanced equation, determine the moles of carbon dioxide and water formed.
Finally, calculate the total combined mass of the products by adding the masses of carbon dioxide and water.
Note: It is important to use accurate molar masses and perform the calculations correctly to obtain the precise answer.