Silver nitrate (AgNO3) reacts with sodium
chloride as indicated by the equation
AgNO3 + NaCl ! AgCl + NaNO3 .
How many grams of NaCl would be re-
quired to react with 511 mL of 0.38 M AgNO3
solution?
Answer in units of grams
Refer to your previous problem.
no
To determine the number of grams of NaCl required to react with the given AgNO3 solution, we can use the following steps:
1. Calculate the number of moles of AgNO3 in the solution:
- Given the volume of AgNO3 solution is 511 mL, we need to convert it to liters by dividing by 1000:
511 mL ÷ 1000 = 0.511 L
- The concentration of AgNO3 is given as 0.38 M, which means there are 0.38 moles of AgNO3 per liter of solution.
- Therefore, the number of moles of AgNO3 in the solution can be calculated as:
Moles of AgNO3 = Volume of solution (in L) × Concentration of AgNO3
Moles of AgNO3 = 0.511 L × 0.38 M = 0.19438 moles
2. According to the balanced chemical equation, the stoichiometry between AgNO3 and NaCl is 1:1. This means that for every 1 mole of AgNO3, we will need 1 mole of NaCl to react.
3. Finally, we can determine the number of grams of NaCl needed using the molar mass of NaCl, which is approximately 58.44 grams/mole:
Grams of NaCl = Moles of NaCl × Molar mass of NaCl
Grams of NaCl = 0.19438 moles × 58.44 g/mol ≈ 11.35 grams
Therefore, approximately 11.35 grams of NaCl would be required to react with 511 mL of 0.38 M AgNO3 solution.