Suppose 0.30 g of ethane is mixed with 0.47 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.
Write the equation and balance it.
Convert 0.30g ethane to moles.
Convert 0.47 g oxygen to moles.
Using the coefficients in the balanced equation, convert moles ethane to moles H2O.
Do the same with moles oxygen.
You will have two answers and both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. The take the smaller number and convert to moles. g = moles x molar mass.
Post your work if you get stuck.
To calculate the maximum mass of water that can be produced, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction, thus limiting the amount of product that can be formed.
To find the limiting reactant, we need to compare the amount of each reactant to their respective stoichiometric coefficients in the balanced chemical equation.
The balanced chemical equation for the combustion of ethane (C2H6) in oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is:
C2H6 + 7/2 O2 → 2 CO2 + 3 H2O
The molar mass of ethane (C2H6) is 30.07 g/mol, and the molar mass of oxygen (O2) is 32.00 g/mol.
First, let's calculate the number of moles for each of the reactants:
Number of moles of ethane (C2H6) = (0.30 g) / (30.07 g/mol) ≈ 0.00997 mol
Number of moles of oxygen (O2) = (0.47 g) / (32.00 g/mol) ≈ 0.01469 mol
Next, we need to determine the mole ratio between the reactants and the product. From the balanced chemical equation, we see that the mole ratio between ethane (C2H6) and water (H2O) is 1:3.
Now, let's calculate the theoretical number of moles of water that can be formed:
Number of moles of water (H2O) = 3 * (0.00997 mol) ≈ 0.02991 mol
Finally, we can calculate the maximum mass of water produced by multiplying the number of moles of water by its molar mass:
Mass of water (H2O) = (0.02991 mol) * (18.02 g/mol) ≈ 0.539 g
Therefore, the maximum mass of water that can be produced by the chemical reaction is approximately 0.539 g.