Hello, Roman,
2 H2(g) + O2(g) → 2 H2O (l)
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An amount of 6 grams of hydrogen is reacted with 64 grams of oxygen.
a. What is the limiting reagent? Justify your answer with a calculation.
We need to convert the grams of each reactant into moles. Do this by dividing the mass by the molar mass (g/(g/mole)).
We see from the balanced equation that we need 2 moles of H2 for every 1 mole of O2. The amount of O2, 2 moles, could react with 4 moles of H2, but we only have 3 moles of H2. Hydrogen is thus the limiting reagent.
b. What is the theoretical yield of the reaction?
The equation says that the nmolar ratio of H2O/H2 is 1:1. If we consume all 3 moles of H2, we would produce 3 moles of water. 3 moles of water can be converyed into grams water by multiplying by it;s molar mass:
(3 moles H2O)*(18 grams/mole) = 54 grams H2O. That is the theorectical yield.
c. How much excess reagent remains after the reaction is complete? (done) 6 grams of H2.
We shouldn't have any H2 left over, since it was the limiting reagent. The 3 moles of H2 will react with 1.5 mole of O2, leaving 0.5 mole unreacted. (0.5 mole O2)*(32g/mole) = 16 grams unreacted oxygen.
d. If 40. grams of water was produced, what is the percent yield of this reaction?
We calculated that 54 grams of water would be produced at 100% yield. If we only got 40 grams, the yield would be (40g/54g)x100%, or 54%. [now is a good time to find some distilled water and bring it up to 100%]
Bob
