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Calculate the enthalpy of the following reaction: 4 b (s) + 3 o2 (g) → 2 b2o3 (s) given the following pertinent information: (a) b2o3 (s) + 3 h2o (g) → 3 o2 (g) + b2h6 (g), δhoa = +2035 kj (b) 2 b (s) + 3 h2 (g) → b2h6 (g), δhob = +36 kj (c) h2 (g) + latex: \frac{1}{2} 1 2 o2 (g) → h2o (l), δhoc = −285 kj (d) h2o (l) → h2o (g), δhod = +44 kj
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Iam hoping to create 5.72 grams of glucose. the plant was given 4.75 liters of co2 and 2.81 g of h20. which reactant was the limiting reagent? how much excess mass did we have of the other reactant?
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