Ask questions and get helpful answers.

A- 50.0 g of H2SO4 react with 75.0 g of NaOH. Identify the limiting and excess reactants. How many grams of Na2SO4 will be formed?

B- A chemist burns 160.0 g of aluminum in 200.0 g of oxygen. She produces 275.0 g of aluminum oxide. Which is the limiting reactant? Which is the excess reactant? How many grams of aluminum oxide should theoretically be produced? What is her percent yield?

1. How many moles are in 125g of H2SO4 ?
2. How many grams are in 1.5 x 1025 molecules of CO2 ?
3. Determine the percentage composition of KMnO4 .
4. What is the empirical formula of a compound that was found to consist of 22.1% aluminum, 25.4% phosphorous, 52.5% oxygen?
5. A compound is found to contain 18.7% lithium, 16.3% carbon and 65.0% oxygen. If the molar mass of the compound is 73.8 g/mol, what is the compound’s molecular formula?

6. How many grams of hydrogen are produced when 25.0 g of zinc reacts with excess Hydrochloric acid?
7. Silver nitrate and Sodium phosphate react in equal amounts of 200.0 g each. How many grams of Silver phosphate are produced?
8. If the reaction above is performed and yields only 155 g of Silver phosphate, what was the percent yield?

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩

1 answer

  1. A- 50.0 g of H2SO4 react with 75.0 g of NaOH. Identify the limiting and excess reactants. How many grams of Na2SO4 will be formed?
    I will work this ONE limiting reagent (LR) problem for you. That should get you started on the others. I work these the long way. Here are the steps.
    1. Write and balance the equation.
    H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
    2. Convert what you have into moles.
    a. mols H2SO4 = grams/molar mass 50/98 = 0.51
    b. mols NaOH = 75/40 = 1.875
    3. Convert EACH of 2a and 2b into any product. You do this by using the coefficients in the balanced equation. I'll choose Na2SO4.
    3a (from 2a.) 0.51 mols H2SO4 x (1 mol Na2SO4/1 mol H2SO4) = 0.51 mols Na2SO4 produced IF we had 50 g H2SO4 and all the NaOH we needed.
    3b (from 2b). 1.875 mol NaOH x (1 mol Na2SO4/2 mol NaOH) = 0.938 mols Na2SO4 IF we had 75 g NaOH and all the H2SO4 we needed.
    4. In LR problems the SMALL number always wins because you can't produce more than the smallest amount; therefore, 0.51 moles of H2SO4 is the LR, NaOH is the excess reagent.
    5. grams Na2SO4 produced = mols Na2SO4 x molar mass Na2SO4 = 0.51 x 142 = ? This is called the theoretical yield (TY)
    If you want percent yield the problem will tell you how much was produced, which I will call AY (for actual yield). Then
    % yield = 100 x (AY/TY) = ?
    Some problems will as for how much of the excess reagent remains. Just run a separate stoichiometry problem using the amount of H2SO4(in this case) to see how much NaOH is consumed and subtract from the initial amount to find the amount remaining.
    Post your work if you get stuck on this or any of the others.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
    👤
    DrBob222

Answer this Question

Related Questions

Still need help?

You can ask a new question or browse existing questions.