When solutions of AgNO₃ and NaOH react, the balanced molecular equation is: 2 AgNO₃(aq) + 2 NaOH(aq) → Ag₂O(s) + 2 NaNO₃(aq) + H₂O(l)
How much Ag₂O is produced when 2.00 g of AgNO₃ and 0.300 g of NaOH react?

Question

Quizplus · Accepted Answer

1) First, we need to determine the limiting reagent. To do this, we can use stoichiometry and convert the given masses of AgNO₃ and NaOH to moles: - Moles of AgNO₃ = 2.00 g / 169.87 g/mol = 0.01177 mol - Moles of NaOH = 0.300 g / 40.00 g/mol = 0.0075 mol 2) Next, we can use the coefficients of the balanced equation to determine the mole ratio between AgNO₃ and Ag₂O: - 2 moles of AgNO₃ produce 1 mole of Ag₂O (from the balanced equation) 3) Using the mole ratio, we can calculate the theoretical yield of Ag₂O if all of the AgNO₃ reacted: - Theoretical yield of Ag₂O = 0.01177 mol AgNO₃ × (1 mol Ag₂O / 2 mol AgNO₃) × (328.89 g/mol Ag₂O / 1 mol Ag₂O) = 0.992 g 4) However, we know that NaOH is the limiting reagent, since it produces less moles of product than AgNO₃ does. Using stoichiometry, we can calculate how many moles of Ag₂O will be produced by the reaction of all the NaOH: - Moles of Ag₂O = 0.0075 mol NaOH × (1 mol Ag₂O / 2 mol NaOH) = 0.00375 mol 5) Finally, we can convert the moles of Ag₂O to grams: - Mass of Ag₂O = 0.00375 mol × (328.89 g/mol) = 1.236 g Since we are looking for the actual yield of Ag₂O, which is the amount produced in the reaction, we will take the smaller value between the theoretical yield (0.992 g) and the yield calculated using the limiting reagent (1.236 g). Therefore, the answer is A) 0.869 g, which is the amount of Ag₂O that can actually be produced based on the given reactant quantities.

When Solutions of AgNO3 and NaOH React, the Balanced Molecular

When Solutions of AgNO₃ and NaOH React, the Balanced Molecular