can you please please help me with these questions I don't know how to do any of them. I know how to calculate moles and grams and mole ratio,etc but I don't understand the limiting excess reactant and yield stuff. Thanks!



1.In the following chemical reaction, how many grams of the excess reactant are left over when 0.45 moles of Al react with 1.25 moles of Cl2 ?
2Al + 3Cl2 → 2AlCl3


2.Using the following equation, determine the % yield from the following reaction if 30.65 g of octane (C8H18 ) react with excess oxygen to produce 81.75 g of CO2

C8 H18 (I) + 25 O2 (g) → 16CO2 + 18H2O

moles of kcl=10.35/74.55=0.138


3.
For the following reaction, determine the amount of AgCl(s) formed when 10.35 g of KCl(aq) react with excess AgNO3 (aq) to produce AgCl(s) and KNO3 (aq) with a 77.15% yield.

AgNO3 + KCl → AgCl(s) + KNO3

1.In the following chemical reaction, how many grams of the excess reactant are left over when 0.45 moles of Al react with 1.25 moles of Cl2 ?

2Al + 3Cl2 → 2AlCl3

LR = limiting reagent. ER - excess reagent. Actually you work three regular stoichiometry problems here. First you want to identify the LR and ER and you do it this way.Use the coefficients in the balanced equation to convert mols Al AND mols Cl2 (in two separate operations) to mols AlCl3.
0.45 mols Al x (2 mols AlCl3/2 mols Al ) = 0.45 mols AlCl3 formed.

1.25 mols Cl2 x (2 mols AlCl3/3 mols Cl2) = 0.83 mols AlCl3.
IN LR problems the smaller number is ALWAYS the mols Product formed and the reagent responsible for that is the LR. In this case that makes Al the LR and Cl2 the ER.
Now that you know Cl2 is the ER, then you need to calculate the amount of Cl2 used. To do that just do the stoichiometry again; i.e., 0.45 mols Al x (3 mols Cl2/2 mols Al) = 0.68 mols Cl2 used.
1.25 mols Cl2 initially.
-0.68 mols Cl2 used
--------
1.25-0.68 = ? mols Cl2 not reacted.


2.Using the following equation, determine the % yield from the following reaction if 30.65 g of octane (C8H18 ) react with excess oxygen to produce 81.75 g of CO2

C8 H18 (I) + 25 O2 (g) → 16CO2 + 18H2O

moles of kcl=10.35/74.55=0.138
I don't know where this 0.138 mols KCl fit so I'll ignore it.

For problem #2, you say you know how to vind mols etc. Do this.
a. Convert grams octane to mols.
b. Use the coefficients in the balanced equation to convert mols octane to mols CO2.
c. Convert mols CO2 to grams CO2. THIS IS THE THEORETICAL YIELD (THE AMOUNT OF STUFF FORMED FOR 100% YIELD). I call this TY for theoretical yield. The actual yield (AY) is 81.75 g
d. %yield = (AY/TY)*100 = ?


3.
For the following reaction, determine the amount of AgCl(s) formed when 10.35 g of KCl(aq) react with excess AgNO3 (aq) to produce AgCl(s) and KNO3 (aq) with a 77.15% yield.

AgNO3 + KCl → AgCl(s) + KNO3

This looks like #2.

thank you so much that makes so much sense :)

I'd be happy to help you with these questions! Let's start by tackling the limiting and excess reactant concept.

1. In order to determine the excess reactant, we first need to identify the limiting reactant. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed. To find the limiting reactant, we compare the moles of each reactant to the mole ratio in the balanced equation.

Given:
moles of Al = 0.45 moles
moles of Cl2 = 1.25 moles

From the balanced equation: 2Al + 3Cl2 → 2AlCl3

The mole ratio of Al to Cl2 is 2:3. So, for every 2 moles of Al, we need 3 moles of Cl2. To determine the limiting reactant, we can calculate the moles of Al and Cl2 required based on this ratio.

moles of Al required = 0.45 moles Al × (3 moles Cl2 / 2 moles Al) = 0.675 moles Cl2
moles of Cl2 required = 1.25 moles Cl2

Since we have more Cl2 than the amount required (1.25 moles), Al is the limiting reactant. This means that all the Al will react, and some Cl2 will be left over as the excess reactant.

To find the excess Cl2, we subtract the moles of Al required from the moles of Cl2 given:
excess Cl2 = moles of Cl2 given - moles of Cl2 required
excess Cl2 = 1.25 moles Cl2 - 0.675 moles Cl2 = 0.575 moles Cl2

Now, to find the grams of the excess Cl2, we can use the molar mass of Cl2:
grams of excess Cl2 = moles of excess Cl2 × molar mass of Cl2

2. To determine the percent yield, we need to compare the actual yield to the theoretical yield. The theoretical yield is the maximum amount of product that can be formed based on the balanced equation and the limiting reactant.

Given:
mass of octane (C8H18) = 30.65 g
mass of CO2 produced = 81.75 g

From the balanced equation: C8H18 + 25O2 → 16CO2 + 18H2O

The molar ratio of C8H18 to CO2 is 1:16. This means that for every mole of C8H18, we produce 16 moles of CO2. To determine the theoretical yield of CO2, we can calculate the moles of C8H18 and then use the mole ratio.

moles of C8H18 = mass of C8H18 / molar mass of C8H18
moles of C8H18 = 30.65 g / molar mass of C8H18

Similarly, we can calculate the moles of CO2 produced using the mass of CO2 and its molar mass.

Theoretical yield of CO2 = moles of C8H18 × (16 moles CO2 / 1 mole C8H18)

To calculate the percent yield, we use the formula:
percent yield = (actual yield / theoretical yield) × 100

Now, let's move on to the third question.

3. To determine the amount of AgCl(s) formed, we need to consider the yield percentage given. The yield percentage represents the ratio of the actual yield to the theoretical yield.

Given:
mass of KCl = 10.35 g
yield percentage = 77.15%

From the balanced equation: AgNO3 + KCl → AgCl(s) + KNO3

First, we need to calculate the moles of KCl using its mass and molar mass.
moles of KCl = mass of KCl / molar mass of KCl

Next, we can use the molar ratio from the balanced equation to determine the moles of AgCl produced and the moles of AgNO3 consumed. The mole ratio is 1:1, so the moles of AgCl formed will be equal to the moles of KCl.

Finally, to find the actual yield of AgCl, we can multiply the moles of AgCl formed by its molar mass.

The actual yield of AgCl = moles of AgCl formed × molar mass of AgCl

To find the amount of AgCl formed considering the yield percentage, we multiply the actual yield by the yield percentage.

Amount of AgCl formed = actual yield of AgCl × (yield percentage / 100)

I hope this guidance helps you understand how to solve these types of questions! Let me know if you need further clarification.