What would be the limiting reagent if 75 grams of C2H2Br3 reacted with 50.0 grams of O2 in the following reaction:
4C2H3Br3 + 11O2-------->8CO2+6H2O+6Br2
To determine the limiting reagent, we need to find the number of moles of each reactant and then compare their mole ratios in the balanced chemical equation.
1. Find the number of moles of C2H3Br3:
75 g C2H3Br3 * (1 mol / 331.64 g) = 0.226 moles C2H3Br3
2. Find the number of moles of O2:
50.0 g O2 * (1 mol / 32.00 g) = 1.56 moles O2
3. Calculate the mole ratio:
C2H3Br3 : O2 = 4 : 11
4. Calculate the number of moles of O2 needed to react with all of the C2H3Br3:
0.226 moles C2H3Br3 * (11 moles O2 / 4 moles C2H3Br3) = 0.62275 moles O2
Comparing the actual moles of O2 (1.56 moles) with the moles needed to react completely (0.62275 moles), we can see that O2 is in excess. Therefore, the limiting reagent in this reaction is C2H3Br3.