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

This is a limiting reagent (LR) problem since amounts are given for BOTH reactants.

AgNO3 + NaBr ==> AgBar + NaNO3

mols AgNO3 = M x L = ?
mols NaBr = M x L = ?

Convert mols AgNO3 to mols AgBr
Convert mols NaBr to mols AgBr.
It is likely the two values will not agree which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for producing that number is the LR.
Now use the smaller value and convert to grams. grams AgBr = mols AgBr x molar mass AgBr.

To determine the mass of AgBr produced when AgNO3 is added to NaBr, we need to consider the balanced chemical equation for the reaction between these two compounds:

AgNO3 (aq) + NaBr (aq) -> AgBr (s) + NaNO3 (aq)

From the balanced equation, we can see that one mole of AgNO3 reacts with one mole of NaBr to produce one mole of AgBr.

To solve this problem, we can use the following steps:

1. Calculate the number of moles of AgNO3:

We multiply the volume (in L) of the AgNO3 solution by its concentration (in mol/L):

moles of AgNO3 = volume of AgNO3 (L) × concentration of AgNO3 (mol/L)

Given that the volume of AgNO3 is 100.0 mL = 0.100 L and the concentration is 0.150 M, we have:

moles of AgNO3 = 0.100 L × 0.150 mol/L = 0.015 mol

2. Calculate the number of moles of NaBr:

Using the same approach, we can calculate the number of moles of NaBr:

moles of NaBr = volume of NaBr (L) × concentration of NaBr (mol/L)

Given that the volume of NaBr is 20.0 mL = 0.0200 L and the concentration is 1.00 M, we have:

moles of NaBr = 0.0200 L × 1.00 mol/L = 0.0200 mol

3. Determine the limiting reagent:

The limiting reagent is the one that gets completely consumed and determines the amount of product formed. To find the limiting reagent, compare the number of moles of AgNO3 and NaBr from the previous steps. The reactant with the smaller number of moles is the limiting reagent.

In this case, AgNO3 has fewer moles (0.015 mol) compared to NaBr (0.0200 mol), so AgNO3 is the limiting reagent.

4. Calculate the moles of AgBr formed:

Since the reaction between AgNO3 and NaBr produces a 1:1 ratio of AgBr to AgNO3, the moles of AgBr formed will also be 0.015 mol.

5. Calculate the mass of AgBr:

The molar mass of AgBr can be used to convert the moles of AgBr to grams:

mass of AgBr = moles of AgBr × molar mass of AgBr

The molar mass of AgBr is the sum of the atomic masses of silver (Ag) and bromine (Br):

molar mass of AgBr = atomic mass of Ag + atomic mass of Br

Using the atomic masses from the periodic table:
molar mass of AgBr = 107.87 g/mol + 79.90 g/mol = 187.77 g/mol

Finally, we can calculate the mass of AgBr:

mass of AgBr = 0.015 mol × 187.77 g/mol = 2.82 g

Therefore, approximately 2.82 grams of AgBr will be produced when 100.0 mL of 0.150 M AgNO3 is added to 20.0 mL of 1.00 M NaBr.