Suppose that 23 mL of 0.6 M K2CrO4(aq) reacts with 13.7 mL of AgNO3(aq) completely.
What mass of NaCl is needed to react completely with 38.5 mL of the same AgNO3 solution?
Answer in units of g
To answer this question, we need to use the stoichiometry of the balanced equation to determine the amount of NaCl needed to react completely with the given volume of AgNO3 solution.
The balanced equation for the reaction between AgNO3 and NaCl is:
AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)
From the given information, we know the volume of the AgNO3 solution (38.5 mL). We need to convert this volume to moles, and then use the stoichiometry of the balanced equation to find the moles of NaCl required. Finally, we can convert the moles of NaCl to grams using its molar mass.
Step 1: Convert the volume of AgNO3 solution to moles.
Given: Volume of AgNO3 solution = 38.5 mL
We need to convert this to liters by dividing by 1000:
38.5 mL ÷ 1000 = 0.0385 L
Now we can calculate the moles of AgNO3 using its molarity:
Molarity of AgNO3(aq) = 0.6 M
Moles of AgNO3 = Molarity × Volume
Moles of AgNO3 = 0.6 M × 0.0385 L = 0.0231 moles
Step 2: Use the stoichiometry of the balanced equation to find moles of NaCl required.
From the balanced equation, we can see that the ratio between AgNO3 and NaCl is 1:1. This means that every mole of AgNO3 reacts with one mole of NaCl.
Therefore, the moles of NaCl required = 0.0231 moles
Step 3: Convert moles of NaCl to grams.
The molar mass of NaCl is 58.44 g/mol.
Mass of NaCl = Moles × Molar mass
Mass of NaCl = 0.0231 moles × 58.44 g/mol = 1.35 g (rounded to two decimal places)
Therefore, the mass of NaCl needed to react completely with 38.5 mL of the AgNO3 solution is 1.35 grams.
To find the mass of NaCl needed to react completely with a given volume of AgNO3 solution, we need to set up and solve a stoichiometry problem.
Step 1: Write the balanced chemical equation for the reaction.
In this case, the reaction is between AgNO3 and NaCl, and it can be written as:
AgNO3 + NaCl → AgCl + NaNO3
Step 2: Determine the stoichiometry of the balanced chemical equation.
From the balanced chemical equation, we can see that the molar ratio between AgNO3 and NaCl is 1:1.
Step 3: Convert the given volume of AgNO3 solution to moles of AgNO3.
Given:
- Volume of AgNO3 solution = 38.5 mL
- We know the molarity of the AgNO3 solution is the same as in the previous question, which is 0.6 M.
To convert the volume of AgNO3 solution to moles, we use the formula:
moles = volume (in liters) x molarity
Converting the volume to liters:
38.5 mL = 38.5 mL * (1 L / 1000 mL) = 0.0385 L
Calculating the moles of AgNO3:
moles of AgNO3 = 0.0385 L x 0.6 mol/L = 0.0231 mol
Step 4: Use the stoichiometry to find the moles of NaCl required.
From the balanced chemical equation, we know that the molar ratio between AgNO3 and NaCl is 1:1. Therefore, the moles of NaCl required will also be 0.0231 mol.
Step 5: Convert moles of NaCl to grams.
To convert the moles of NaCl to grams, we need to know the molar mass of NaCl. The molar mass of NaCl is 58.44 g/mol.
Calculating the mass of NaCl:
mass of NaCl = moles of NaCl x molar mass of NaCl
mass of NaCl = 0.0231 mol x 58.44 g/mol = 1.35 g
Therefore, 1.35 grams of NaCl is needed to react completely with 38.5 mL of the AgNO3 solution.