if 2.70 moles of C2H2 is reacted with 6.30moles of O2, calculate the moles of water produced
To calculate the moles of water produced, we need to determine the balanced equation for the reaction between C2H2 (acetylene) and O2 (oxygen) first.
The balanced equation for the combustion of acetylene is:
C2H2 + O2 --> CO2 + H2O
From the balanced equation, we see that for every 2 moles of acetylene (C2H2) reacted, we need 5 moles of oxygen (O2) to produce 2 moles of water (H2O).
Given that we have 2.70 moles of C2H2 and 6.30 moles of O2, we can determine the limiting reactant (the reactant that is completely consumed) to find the moles of water produced.
First, we calculate the moles of C2H2 needed to react with O2:
2.70 moles of C2H2 * (5 moles of O2 / 2 moles of C2H2) = 6.75 moles of O2 needed
Since we only have 6.30 moles of O2, we conclude that O2 is the limiting reactant.
To determine the moles of water produced, we use the stoichiometry from the balanced equation:
6.30 moles of O2 * (2 moles of H2O / 5 moles of O2) = 2.52 moles of H2O
Therefore, the moles of water produced is 2.52 moles.