1. Iron oxide and carbon monoxide react according to the following equation . Identify which reactant is limiting and which is excess if 43.5 grams of Fe2O3 reacts with 43.5 grams of CO
Fe2O3 + 3CO2 --> 3CO2 + 2Fe
2. Calculate the percent yield of Carbon dioxide if 20.0 grams of it are produced when 19.5 grams of butane are burned in excess O2 according to the following equation
2C4H10 + 13O2 --> 8CO2 + 10 H2O
3. Calculate the percent yield of copper nitrate if 356 grams of it are produced when 202 grams of copper nitrate are reacted according to the following equation
3Cu + 8HNO3 --> 3Cu (NO3)2 + 2NO + 4H2O
4. When 150 grams of ZnS are burned in excess oxygen 0.685 grams of ZnO are produced according to the following equation. What is the percent yield for this reaction.
2ZnS + 3O2 --> 2ZnO + 2SO2
5. What is the actual yield in grams of oxygen if you start with 100 grams of H2O2 ? The reaction proceeds according to the following equation
2H2O2 --> 2H2O + O2
I'm don't really know how to solve any of these problems. Can you show me how to do each of these prioblems ? Please I really need help
a. calculate the moles of reactant/products for the givens.
b. use the mole ratio in the balanced equation to calcuate the unknowns.
c. reconvert to mass.
Example: Number 3
3Cu + 8HNO3 --> 3Cu (NO3)2 + 2NO + 4H2O
given 356 grams Cu(NO3)3, or 356/187.6 moles (which is equal to 1.90 moles
given 202 grams Cu (I assume you erred on the typed problem, you meant 202 g Cu). Moles Cu: 202/63.5 =3.2 moles
a. Notice the mole ratio in the balanced equation, it is 3:3 (or 1 to 1).
So in a perfect world, you should have gotten 3.2 moles copper nitrate from the 3.2 moles Cu.
Percent yield: 1.90/3.2= you do it.
Now, on the problems with limiting reactants.
look at your moles given. Calculate the moles for the other reactant from the ratios. If you have more than needed, then the given reactant is limiting.
Of course! I can help you solve each of these problems step by step. Let's start with problem 1:
1. To determine which reactant is limiting and which is in excess, we need to compare the amount of each reactant to the stoichiometric ratio in the balanced equation.
Given:
Mass of Fe2O3 = 43.5 grams
Mass of CO = 43.5 grams
Balanced equation: Fe2O3 + 3CO --> 3CO2 + 2Fe
a) Calculate the number of moles for each reactant:
Molar mass of Fe2O3 = 159.69 g/mol
Molar mass of CO = 28.01 g/mol
Number of moles of Fe2O3 = Mass / Molar mass = 43.5 g / 159.69 g/mol
Number of moles of CO = Mass / Molar mass = 43.5 g / 28.01 g/mol
b) Compare the moles of each reactant to the stoichiometric ratio in the balanced equation. From the balanced equation, we can see that the ratio of Fe2O3 to CO is 1:3.
Since 1 mole of Fe2O3 reacts with 3 moles of CO, we need to determine which reactant has the smallest number of moles.
c) Calculate the mole ratio for each reactant:
Moles of Fe2O3 / Stoichiometric coefficient of Fe2O3 = 43.5 g / (159.69 g/mol) = 0.2723 mol
Moles of CO / Stoichiometric coefficient of CO = 43.5 g / (28.01 g/mol) = 1.553 mol
d) Compare the mole ratios:
Fe2O3:CO = 0.2723 mol : 1.553 mol
Since Fe2O3 has the smaller mole ratio, it is the limiting reactant, and CO is in excess.
Moving on to problem 2:
2. To calculate the percent yield of carbon dioxide, we need to compare the actual yield of CO2 to the theoretical yield, and then express it as a percentage.
Given:
Mass of CO2 produced = 20.0 grams
Mass of butane reacted = 19.5 grams
Balanced equation: 2C4H10 + 13O2 --> 8CO2 + 10H2O
a) Calculate the number of moles for each reactant and product:
Molar mass of CO2 = 44.01 g/mol
Molar mass of butane (C4H10) = 58.12 g/mol
Number of moles of CO2 = Mass / Molar mass = 20.0 g / 44.01 g/mol
Number of moles of butane = Mass / Molar mass = 19.5 g / 58.12 g/mol
b) Determine the stoichiometric ratio between CO2 and butane from the balanced equation. From the balanced equation, we can see that the ratio of butane to CO2 is 2:8 (which simplifies to 1:4).
c) Calculate the theoretical yield of CO2:
Theoretical yield of CO2 = Moles of butane * (Moles of CO2 / Moles of butane) * Molar mass of CO2
= (19.5 g / 58.12 g/mol) * (8 mol CO2 / 2 mol butane) * 44.01 g/mol
d) Calculate the percent yield of CO2:
Percent yield = (Actual yield of CO2 / Theoretical yield of CO2) * 100
= (20.0 g / Theoretical yield) * 100
For problem 3, 4, and 5, please provide more specific information such as the balanced equations and any given masses.