When MoO3 and Zn are heated together they react
3Zn(s)+2MoO3(s) --> Mo2O3(s)+3ZnO(s)
What mass of ZnO is formed when 20.0 grams of MoO3 is reacted with 10.0 grams of Zn?
Do you know how to do limiting reagent problems. This is one? What do you know about them?
I know how to convert moles to mass, atoms, volume etc. and I can find the limiting reagent and the percentage yield of some basic problems.
This is a limiting reagent problem; you know that because amounts for BOTH reactants are given.
Find the limiting reagent, then convert that, using the coefficients in the balanced equation, to mols of the product. Then convert to g. g = mols x molar mass.
Thank you so much!
To find the mass of ZnO formed when 20.0 grams of MoO3 and 10.0 grams of Zn are reacted, we first need to determine which reactant is limiting and which is in excess.
To do this, we will use the concept of stoichiometry and the given balanced chemical equation:
3Zn(s) + 2MoO3(s) → Mo2O3(s) + 3ZnO(s)
First, let's calculate the number of moles for each reactant.
1. Calculate the number of moles of MoO3:
Given mass of MoO3 = 20.0 grams
Molar mass of MoO3 = atomic mass of Mo + 3 * atomic mass of O
= (95.95 g/mol) + 3 * (16.00 g/mol)
= 95.95 g/mol + 48.00 g/mol
= 143.95 g/mol
Number of moles of MoO3 = mass of MoO3 / molar mass of MoO3
= 20.0 g / 143.95 g/mol
≈ 0.139 moles
2. Calculate the number of moles of Zn:
Given mass of Zn = 10.0 grams
Molar mass of Zn = 65.38 g/mol
Number of moles of Zn = mass of Zn / molar mass of Zn
= 10.0 g / 65.38 g/mol
≈ 0.153 moles
Next, we need to determine which reactant is limiting. The reactant that is completely consumed determines the maximum amount of product that can be formed. We can compare the mole ratios of the reactants from the balanced equation to determine the limiting reactant.
From the balanced equation, we can see that the mole ratio between MoO3 and Zn is 2:3. This means that for every 2 moles of MoO3, we need 3 moles of Zn to react completely.
To determine the limiting reactant:
Divide the number of moles of each reactant by their respective stoichiometric coefficients:
Moles of MoO3 / coefficient = 0.139 moles / 2
≈ 0.070 moles
Moles of Zn / coefficient = 0.153 moles / 3
≈ 0.051 moles
Based on the calculations, Zn is the limiting reactant because it produces a smaller number of moles than MoO3.
Now that we know Zn is the limiting reactant, we can calculate the mass of ZnO formed. We'll assume the reaction goes to completion, meaning all of the limiting reactant is consumed.
To determine the number of moles of ZnO formed, use the mole ratio between ZnO and Zn from the balanced equation, which is 1:3.
Number of moles of ZnO = Moles of Zn * coefficient
= 0.051 moles * 3
≈ 0.153 moles
Finally, to find the mass of ZnO:
Mass of ZnO = Number of moles of ZnO * molar mass of ZnO
≈ 0.153 moles * (81.38 g/mol) (molar mass of ZnO)
≈ 12.429 g
Therefore, approximately 12.429 grams of ZnO is formed when 20.0 grams of MoO3 is reacted with 10.0 grams of Zn.