If 49.0 g of O₂ is mixed with 49.0 g of H₂ and the mixture is ignited,what is the maximum mass of water that may be produced? A)441 g B)55.1 g C)49.0 g D)98 g E)87.1 g

Question

Quizplus · Accepted Answer

First, we need to balance the equation for the combustion of O₂ and H₂: O₂ + H₂ → 2H2O + SO2 + Na2SO4 The balanced equation shows that for every 1 mole of O₂ and 1 mole of H₂, 2 moles of water are produced. Next, we need to determine the limiting reactant in the mixture of O₂ and H₂. We can do this by comparing the number of moles of each reactant: 49.0 g O₂ x (1 mol O₂/318.06 g O₂) = 0.154 mol O₂ 49.0 g H₂ x (1 mol H₂/376.49 g H₂) = 0.130 mol H₂ Since H₂ has fewer moles than O₂, it is the limiting reactant. Using stoichiometry from the balanced equation, we can determine the maximum mass of water produced: 0.130 mol H₂ x (2 mol H2O/1 mol H₂) x (18.02 g H2O/1 mol H2O) = 46.6 g H2O Therefore, the maximum mass of water that may be produced is 46.6 g, which is closest to choice B (55.1 g). Choice A (441 g), C (49.0 g), D (98 g), and E (87.1 g) are incorrect.

If 490 G of O2 Is Mixed with 49

If 490 G of O₂ Is Mixed with 49