Final answer:
The grams of C₂H₂ present after the reaction is 10.5 grams, the grams of O₂ present after the reaction is 10.5 grams, the grams of CO₂ present after the reaction is 53.328 grams, and the grams of H₂O present after the reaction is 29.088 grams.
Step-by-step explanation:
According to the given balanced equation: C₂H₂(g) +50₂(g) → 3CO₂(g) + 4H₂O(g)
We can calculate the number of moles of C₂H₂ and O₂ using their masses. The molar mass of C₂H₂ is 26 g/mol and that of O₂ is 32 g/mol. Therefore, the number of moles of C₂H₂ is 10.5 g / 26 g/mol = 0.404 moles. The number of moles of O₂ is 10.5 g / 32 g/mol = 0.328 moles.
Since the reaction stoichiometry is 1:3 for C₂H₂ to CO₂ and 1:4 for C₂H₂ to H₂O, the number of moles of CO₂ produced is 3 times the number of moles of C₂H₂, which is 3 * 0.404 = 1.212 moles. Similarly, the number of moles of H₂O produced is 4 times the number of moles of C₂H₂, which is 4 * 0.404 = 1.616 moles.
To calculate the grams of each compound present after the reaction, we can use the molar masses. The molar mass of CO₂ is 44 g/mol and that of H₂O is 18 g/mol. Therefore, the grams of C₂H₂ present after the reaction is 0.404 moles * 26 g/mol = 10.5 grams. The grams of O₂ present after the reaction is 0.328 moles * 32 g/mol = 10.5 grams. The grams of CO₂ present after the reaction is 1.212 moles * 44 g/mol = 53.328 grams. The grams of H₂O present after the reaction is 1.616 moles * 18 g/mol = 29.088 grams.