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| 1. |
. If the reaction of 40.8 grams of C6H6O3 with oxygen produces a 39.0% yield, how many grams of water would be produced? |
| Answer» EXPLANATION:c6h6o3 + o2 = 6co2+3h2ohere o2 is excess so c6h6o3 is LIMITING reagentfind limiting RATIO of c6h6o3 = moles / stichiometric coefficient moles of c6h6o3 =0.317limiting ratio = 0.317moles of H2O produced when 100 percent yeildlimiting ratio*stichiometric coefficient of h2o0.317*30.951moles of h2o produced when 80 percent yeild0.8*0.9510.761moles of h2o = 0.761w/18=0.761w=13.71 GM | |