We appreciate your visit to Consider the following reaction text C 3 text H 8 5 text O 2 rightarrow 3 text CO 2 4 text H 2 text O. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
if you begin with 37.1 grams of each reactant, you can produce 60.71 grams of water vapor.
To determine the grams of water vapor produced, we need to perform a stoichiometric calculation based on the given balanced equation.
Step 1: Calculate the number of moles of C3H8: Molar mass of C3H8 = (3 × 12.01 g/mol) + (8 × 1.01 g/mol) = 44.11 g/mol Number of moles of C3H8 = 37.1 g / 44.11 g/mol = 0.841 mol
Step 2: Calculate the number of moles of O2: Molar mass of O2 = (2 × 16.00 g/mol) = 32.00 g/mol Number of moles of O2 = 37.1 g / 32.00 g/mol = 1.16 mol
Step 3: Determine the limiting reactant: From the balanced equation, the stoichiometric ratio between C3H8 and H2O is 4:4. Comparing the moles of C3H8 and O2, we see that 1 mole of C3H8 produces 4 moles of H2O, while 1 mole of O2 produces 4 moles of H2O. Therefore, the limiting reactant is C3H8 since it is fully consumed in the reaction.
Step 4: Calculate the grams of water vapor produced: Molar mass of H2O = (2 × 1.01 g/mol) + (16.00 g/mol) = 18.02 g/mol Grams of H2O produced = 0.841 mol × 4 mol × 18.02 g/mol = 60.71 g
Therefore, if you begin with 37.1 grams of each reactant, you can produce 60.71 grams of water vapor.
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