what mass of solid AgBr is produced when 100.0 mL of 0.150 M AgNO3 is added to 20.0mL of 1.00 MNaBr?

i keep getting the answer 3.75 g however when i checked in the back of the book it gives me 2.82 g.. can you please explain the universal steps for me

The universal steps:

1) balanced equation
AgNO3 + NaBr >> NaNO3 (aq) + AgBr (s)

balanced.
So, one mole of AgBr is produced for each mole of silver nitrate and sodium bromide.

2) number moles you started with:
AgNO3= .1*.150= .0150moles

NaBr= .020*1= .0200 moles

3) determining the limiting reactant: Since the reactants react on a one to one basis, the AgNO3 is the limiting reactant. You have .0150 moles of that.

4) determine the moles of product:
.0150 moles of AgBr, from the balanced equation

5) mass produced: .0150*molmassAgBr=.0150*188=2.82 grams

2.82g

To find the mass of solid AgBr produced, we need to determine which reactant is limiting and calculate the amount of AgBr produced from that reactant.

Let's go step-by-step to solve the problem:

Step 1: Write and balance the chemical equation:
AgNO3 + NaBr -> AgBr + NaNO3

Step 2: Calculate the moles of AgNO3 and NaBr:

Moles of AgNO3 = volume (L) x molarity (mol/L)
Moles of AgNO3 = 0.100 L x 0.150 mol/L
Moles of AgNO3 = 0.015 mol

Moles of NaBr = volume (L) x molarity (mol/L)
Moles of NaBr = 0.0200 L x 1.00 mol/L
Moles of NaBr = 0.0200 mol

Step 3: Determine the limiting reactant:
The reactant that produces the smallest number of moles is the limiting reactant.

In the balanced chemical equation, the stoichiometric ratio between AgNO3 and AgBr is 1:1. Therefore, 0.015 mol of AgNO3 produces 0.015 mol of AgBr.

In the balanced chemical equation, the stoichiometric ratio between NaBr and AgBr is also 1:1. Therefore, 0.0200 mol of NaBr produces 0.0200 mol of AgBr.

Since 0.015 mol of AgNO3 produces a smaller amount of AgBr (0.015 mol), AgNO3 is the limiting reactant.

Step 4: Calculate the moles of AgBr produced:
Since AgNO3 is the limiting reactant, we know that 0.015 mol of AgBr is produced.

Step 5: Convert moles to grams:
To convert moles of AgBr to grams, we need to use the molar mass of AgBr.

The molar mass of AgBr is:
AgBr = (107.87 g/mol) + (79.90 g/mol)
AgBr = 187.77 g/mol

Mass of AgBr = moles x molar mass
Mass of AgBr = 0.015 mol x 187.77 g/mol
Mass of AgBr = 2.817 g

Therefore, the mass of AgBr produced is 2.817 g.

The book's answer of 2.82 g is correct. Your initial answer of 3.75 g seems to have been calculated incorrectly.

To determine the mass of solid AgBr produced, you need to use stoichiometry, which involves converting the given amounts of reactants (AgNO3 and NaBr) to the desired product (AgBr).

Here are the universal steps to solve this problem:

Step 1: Write the balanced chemical equation.
The balanced equation for the reaction between AgNO3 and NaBr is:
AgNO3 + NaBr -> AgBr + NaNO3

Step 2: Determine the mole ratios.
From the balanced equation, you can see that 1 mole of AgNO3 reacts with 1 mole of NaBr to produce 1 mole of AgBr.

Step 3: Calculate the moles of each reactant.
To find the moles of AgNO3, use the equation:
moles of AgNO3 = volume (in liters) x concentration (in Molarity)

Given:
Volume of AgNO3 = 100.0 mL = 100.0/1000 = 0.100 L
Concentration of AgNO3 = 0.150 M

moles of AgNO3 = 0.100 L x 0.150 M = 0.015 moles of AgNO3

Similarly, for NaBr:
Volume of NaBr = 20.0 mL = 20.0/1000 = 0.020 L
Concentration of NaBr = 1.00 M

moles of NaBr = 0.020 L x 1.00 M = 0.020 moles of NaBr

Step 4: Determine the limiting reactant.
To find the limiting reactant, compare the moles of each reactant. The reactant that produces the smaller amount of product is the limiting reactant.

From step 3, we have:
moles of AgNO3 = 0.015 moles
moles of NaBr = 0.020 moles

In this case, the limiting reactant is AgNO3 since it produces fewer moles of AgBr.

Step 5: Calculate the moles of AgBr produced.
Since AgNO3 is the limiting reactant, the moles of AgBr produced will be the same as the moles of AgNO3 reacted.

moles of AgBr = 0.015 moles

Step 6: Convert moles of AgBr to grams.
To convert moles to grams, you need to know the molar mass of AgBr. The molar mass of Ag is 107.87 g/mol, and Br is 79.90 g/mol.

Molar mass of AgBr = 107.87 g/mol + 79.90 g/mol = 187.77 g/mol

mass of AgBr = moles of AgBr x molar mass of AgBr

mass of AgBr = 0.015 moles x 187.77 g/mol = 2.82 g

Therefore, the correct answer is 2.82 grams, not 3.75 grams as you initially calculated.

Keep in mind that the molar mass and stoichiometry are crucial in these calculations. Make sure to check your units and significant figures along the way to ensure accurate results.