Asked by Jeff D
What is the limiting reactant in this experiment? Assume you used 1.0 g of the copper(II) sulfate pentahydrate and 7. 5 mL of 6.0 M NH3. Show your calculations.
Its been a while need some help, thanks!
Its been a while need some help, thanks!
Answers
Answered by
DrBob222
And what is the experiment? What is the reaction? Is this it?
CuSO4.5H2O + 4NH3 --> [Cu(NH3)4]SO4.5H2O
mols CuSO4.5H2O = grams/molar mass = ?
mols NH3 = M x L = ?
Now convert mols each to mols product using the coefficients in the balanced equation. It's done this way.
mols product = mols CuSO4.5H2O x (1 mol product/1 mol CuSO4.5H2O) = ?
mols product = mols NH3 x (1 mol product/4 mols NH3) = ?
The smaller number is the amount of product formed and the reagent responsible for that is the limiting reagent.
CuSO4.5H2O + 4NH3 --> [Cu(NH3)4]SO4.5H2O
mols CuSO4.5H2O = grams/molar mass = ?
mols NH3 = M x L = ?
Now convert mols each to mols product using the coefficients in the balanced equation. It's done this way.
mols product = mols CuSO4.5H2O x (1 mol product/1 mol CuSO4.5H2O) = ?
mols product = mols NH3 x (1 mol product/4 mols NH3) = ?
The smaller number is the amount of product formed and the reagent responsible for that is the limiting reagent.
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