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

See your post above.

How many moles of CaO are needed to react with excess water to produce 370g of calcium hydroxide?

Of course, I can help you with those questions! Let's tackle them one by one and I'll explain the concepts of limiting reactant and percent yield along the way.

1. In order to determine the grams of excess reactant left over, you first need to identify which reactant is the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction, determining the maximum amount of product that can be formed.

To find the limiting reactant, you can use the mole ratio from the balanced equation. In this case, the balanced equation is:

2Al + 3Cl2 → 2AlCl3

First, calculate the moles of Al you have: 0.45 moles.
Then, calculate the moles of Cl2 you have: 1.25 moles.

To determine the limiting reactant, compare the moles of each reactant to the stoichiometric ratio in the balanced equation. In this case, the stoichiometric ratio is 2:3 (Al:Cl2).

For Al: 0.45 moles × (3 moles Cl2 / 2 moles Al) = 0.675 moles Cl2 consumed
For Cl2: 1.25 moles

Since 0.675 moles Cl2 is more than the 1.25 moles you have, Cl2 is the limiting reactant. Therefore, Al is the excess reactant.

To find the grams of excess reactant left over, you can use the molar mass of Al and the moles of Al you have. Multiply the moles of Al by the molar mass of Al:

0.45 moles Al × (26.98 g/mol Al) = 12.141 g Al (rounded to three decimal places)

Therefore, approximately 12.141 grams of Al are left over.

2. To determine the percent yield, you need to compare the actual yield (given in the question) and the theoretical yield (calculated from stoichiometry).

The balanced equation for this reaction is:

C8H18 + 25O2 → 16CO2 + 18H2O

First, find the moles of octane (C8H18) by dividing the given mass by the molar mass:

30.65 g C8H18 / (114.22 g/mol C8H18) = 0.268 moles C8H18

From the balanced equation, you can see that the stoichiometric ratio between C8H18 and CO2 is 1:16. Therefore, the theoretical yield of CO2 can be calculated by multiplying the moles of C8H18 by the stoichiometric factor:

0.268 moles C8H18 × (16 moles CO2 / 1 mole C8H18) = 4.288 moles CO2

To calculate the theoretical yield of CO2 in grams, multiply the moles of CO2 by the molar mass of CO2:

4.288 moles CO2 × (44.01 g/mol CO2) = 188.71 g CO2 (rounded to two decimal places)

Next, divide the actual yield (81.75 g CO2) by the theoretical yield (188.71 g CO2) and multiply by 100 to get the percent yield:

(81.75 g CO2 / 188.71 g CO2) × 100 = 43.33% yield (rounded to two decimal places)

Therefore, the percent yield of CO2 is approximately 43.33%.

3. To determine the amount of AgCl(s) formed, you need to consider the percent yield given in the question.

The balanced equation for this reaction is:

AgNO3 + KCl → AgCl(s) + KNO3

First, calculate the moles of KCl using the given mass and the molar mass of KCl:

10.35 g KCl / (74.55 g/mol KCl) = 0.139 moles KCl

Since the reaction is specified to have a 77.15% yield, you can calculate the theoretical yield of AgCl by dividing the actual yield by the percent yield:

0.139 moles KCl / 0.7715 (percent yield) = 0.180 moles AgCl (rounded to three decimal places)

To determine the mass of AgCl formed, multiply the moles of AgCl by the molar mass of AgCl:

0.180 moles AgCl × (143.32 g/mol AgCl) = 25.81 g AgCl (rounded to two decimal places)

Therefore, approximately 25.81 grams of AgCl are formed.