how many grams of a precipitate will form when you mix 45 ml of a 3.5M silver nitrate solution with 45 ml of a 5.4 M magnesium chloride solution

i sorted the problem but have no clue how to start

2AgNO3 + MgCl2 ==> 2AgCl + Mg(NO3)2

How many moles AgNO3 do you have? M x L = moles.
How many moles MgCl2 do you have? M x L = moles.
Calculate moles AgCl that will form (using the equation, of course), then convert to grams.

you have 157.5 mmol of AgNO3 and 243 mmol of MgCl2. From the balanced chemical equation, 1 mol of MgCl2 would consume 2 moles of AgNO3. It would then require 315 mmol of AgNO3 to completely consume the MgCl2. Therefore, AgNO3 is the limiting reactant and it would dictate how much AgCl would precipitate out.

Since the ratio of AgCl to AgNO3 is 1:1, multiplying 157.5 mmol with the molecular weight of AgCl (in mg/mmol) would give 26.755g AgCl.

To determine the number of grams of precipitate that will form when mixing silver nitrate and magnesium chloride solutions, you need to use stoichiometry and the concept of limiting reagents.

Here's how you can approach the problem:

Step 1: Write the balanced equation for the reaction between silver nitrate (AgNO3) and magnesium chloride (MgCl2). The balanced equation is:

AgNO3 + MgCl2 → AgCl + Mg(NO3)2

Step 2: Calculate the number of moles of silver nitrate (AgNO3) and magnesium chloride (MgCl2) in the solutions.

Given:
Volume of silver nitrate solution (V1) = 45 mL = 0.045 L
Molarity of silver nitrate solution (M1) = 3.5 M

Number of moles of AgNO3 = Molarity × Volume
Number of moles of AgNO3 = 3.5 mol/L × 0.045 L = 0.1575 moles

Repeat the same calculation for magnesium chloride:
Volume of magnesium chloride solution (V2) = 45 mL = 0.045 L
Molarity of magnesium chloride solution (M2) = 5.4 M

Number of moles of MgCl2 = Molarity × Volume
Number of moles of MgCl2 = 5.4 mol/L × 0.045 L = 0.243 moles

Step 3: Determine the limiting reagent.
Compare the number of moles of each reactant (AgNO3 and MgCl2) to find the limiting reagent. The limiting reagent is the one that is completely consumed in the reaction, limiting the amount of product formed.

From the balanced equation, you can see that 1 mole of AgNO3 reacts with 2 moles of MgCl2 to produce 1 mole of AgCl. Therefore, the moles of AgNO3 and MgCl2 are in a 1:2 ratio.

Since there are 0.1575 moles of AgNO3 and 0.243 moles of MgCl2, AgNO3 is the limiting reagent because it requires twice as many moles of MgCl2 to react completely.

Step 4: Determine the moles and grams of precipitate formed.
From the balanced equation, you know that 1 mole of AgNO3 reacts to form 1 mole of AgCl.

Since the limiting reagent is AgNO3, the number of moles of AgCl formed will be equal to the number of moles of AgNO3.

Number of moles of AgCl = 0.1575 moles

To convert moles into grams, you need the molar mass of AgCl. The molar mass of AgCl can be calculated by adding the atomic masses of silver (Ag) and chlorine (Cl).

Molar mass of AgCl = atomic mass of Ag + atomic mass of Cl
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol

Number of grams of AgCl = Number of moles × Molar mass
Number of grams of AgCl = 0.1575 moles × 143.32 g/mol = 22.6 grams

Therefore, approximately 22.6 grams of AgCl precipitate will form when you mix 45 mL of a 3.5 M silver nitrate solution with 45 mL of a 5.4 M magnesium chloride solution.