Final answer:
After the reaction, there will be 0.384 grams of C₂H₂, 0.313 grams of O₂, 176.04 grams of CO₂, and 90.10 grams of H₂O.
Step-by-step explanation:
According to the given balanced chemical equation:
2 C₂H₂ (g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(g)
To find how many grams of C₂H₂, O₂, CO₂, and H₂O are present after the reaction, we need to calculate the moles of each substance first.
Using the molar mass of C₂H₂ (26.04 g/mol) and O₂ (32.00 g/mol), we can calculate the moles of C₂H₂ and O₂ by dividing their masses by their molar masses:
C₂H₂: 10 g / 26.04 g/mol = 0.384 moles
O₂: 10 g / 32.00 g/mol = 0.313 moles
According to the stoichiometry of the balanced equation, we can determine the moles of CO₂ and H₂O produced:
CO₂: 8 moles of CO₂ are produced for every 2 moles of C₂H₂, so
8/2 = 4 moles of CO₂ are produced
H₂O: 10 moles of H₂O are produced for every 2 moles of C₂H₂, so
10/2 = 5 moles of H₂O are produced
Finally, we can calculate the masses of CO₂ and H₂O using their molar masses:
CO₂: 4 moles x 44.01 g/mol = 176.04 g
H₂O: 5 moles x 18.02 g/mol = 90.10 g
Therefore, after the reaction, there will be 0.384 grams of C₂H₂, 0.313 grams of O₂, 176.04 grams of CO₂, and 90.10 grams of H₂O.