A mixture of 10cm3 of oxygen and 50cm3 of hydrogen is sparked continuously. Calculate volume of non limiting reagent left after reaction completion:
Correct answer: D. 30cm3
- A. 10cm3
- B. 15cm3
- C. 20cm3
- D. 30cm3
Explanation
We'll begin by writing the balanced equation for the reaction. This is given below: O2 + 2H2 -> 2H2O From the balanced equation above, we can say that: 1 cm³ of O2 reacted with 2 cm³ of H2 to produce 2 cm³ of H2O. Next, we shall determine the excess reactant. This can be obtained as follow: From the balanced equation above, 1 cm³ of O2 reacted with 2 cm³ of H2. Therefore, 10 cm³ of O2 will react with = (10 × 2)/1 = 10 × 2 = 20 cm³ of H2. From the calculations made above, we can see that only 20 cm³ out of 50 cm³ of H2 given is needed to react completely with 10 cm³ of O2. Therefore, O2 is the limiting reactant and H2 is the excess reactant (non limiting reactant). Finally, we shall determine the volume of the non limiting reactant (excess react) that is remaining after the reaction. This can be obtained as follow: Volume of non limiting reactant (H2) = 50 cm³ Volume of non limiting reactant (H2) that reacted = 20 cm³ Volume of non limiting reactant (H2) remaining = (Volume of non limiting reactant) - (Volume of non limiting reactant that reacted) Volume of non limiting reactant (H2) remaining = 50 - 20 = 30 cm³
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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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