If 15 grams of copper (11) chloride react with 20 grams of sodium nitrate, how much sodium chloride can be formed? Hint: you will need two stochiometry problems to determine which reactants limit the amount of NaCl produced?
How do you figure out the 2 problems you will need? This is on a worksheet and this answer determines the rest of the answers
13.0 grams
CuCl2 + 2NaNO3 ==>Cu(NO3)2 + 2NaCl
1. Convert 15 g CuCl2 to moles. moles = grams/molar mass. Convert 20 g NaNO3 to moles.
2a. Using the coefficients in the balanced equation, convert moles CuCl2 to moles NaCl.
2b. Using the coefficients in the balanced equation, convert moles NaNO3 to moles NaCl.
2c. The answer to 2a and 2b likely will be different. Obviously, both can't be correct.
2c. In limiting reagent problems, such as this one, the correct answer is ALWAYS the smaller value and that gives the limiting reagent.
3. Now convert the moles NaCl found in 2c to grams. g = moles x molar mass.
To determine the two stoichiometry problems needed to solve this question, you need to compare the molar ratios of the reactants involved in the chemical equation. Once you determine the limiting reactant, you can then determine the maximum amount of sodium chloride that can be formed.
Here's the step-by-step process to figure out the two stoichiometry problems:
1. Write the balanced chemical equation for the reaction between copper (II) chloride and sodium nitrate:
2CuCl2 + 2NaNO3 -> 2Cu(NO3)2 + 4NaCl
2. Calculate the molar mass of each compound:
- Copper (II) chloride (CuCl2) = 63.55 g/mol + (2 * 35.45 g/mol) = 134.40 g/mol
- Sodium nitrate (NaNO3) = 22.99 g/mol + 14.01 g/mol + (3 * 16.00 g/mol) = 85.00 g/mol
3. Convert the given masses of Reactant A (copper (II) chloride) and Reactant B (sodium nitrate) into moles:
- Moles of copper (II) chloride = mass / molar mass = 15 g / 134.40 g/mol = 0.1117 mol
- Moles of sodium nitrate = mass / molar mass = 20 g / 85.00 g/mol = 0.2353 mol
4. Determine the stoichiometry ratio between copper (II) chloride and sodium chloride by comparing the coefficients in the balanced equation. The ratio is 1:2 (1 mole of copper (II) chloride produces 2 moles of sodium chloride).
5. Calculate the maximum moles of sodium chloride that can be formed from Reactant A:
- Moles of sodium chloride from copper (II) chloride = moles of copper (II) chloride * (2 moles of sodium chloride / 1 mole of copper (II) chloride)
- Moles of sodium chloride from copper (II) chloride = 0.1117 mol * (2 moles / 1 mole) = 0.2234 mol
6. Determine the stoichiometry ratio between sodium nitrate and sodium chloride by comparing the coefficients in the balanced equation. The ratio is 2:4 (2 moles of sodium nitrate produce 4 moles of sodium chloride).
7. Calculate the maximum moles of sodium chloride that can be formed from Reactant B:
- Moles of sodium chloride from sodium nitrate = moles of sodium nitrate * (4 moles of sodium chloride / 2 moles of sodium nitrate)
- Moles of sodium chloride from sodium nitrate = 0.2353 mol * (4 moles / 2 moles) = 0.4706 mol
8. Compare the two calculated values of moles of sodium chloride. The smaller value represents the limiting reactant, as it determines the maximum amount of product that can be formed.
9. Use the limiting reactant to calculate the maximum mass of sodium chloride that can be formed:
- Mass of sodium chloride = moles of limiting reactant * molar mass of sodium chloride
- Mass of sodium chloride = 0.2234 mol * (22.99 g/mol + 35.45 g/mol) = 16.83 g
Therefore, the maximum amount of sodium chloride that can be formed is 16.83 grams.
To determine the two stoichiometry problems needed to solve this question, we need to compare the amounts of copper (11) chloride and sodium nitrate and identify which reactant will limit the amount of sodium chloride produced.
Step 1: Write the balanced chemical equation.
The balanced chemical equation for the reaction between copper (11) chloride and sodium nitrate is:
2 CuCl2 + 2 NaNO3 --> 2 Cu(NO3)2 + 4 NaCl
Step 2: Calculate the molar masses.
The molar mass of copper (11) chloride (CuCl2) is 134.45 g/mol (63.55 g/mol for copper + 2 * 35.45 g/mol for chlorine).
The molar mass of sodium nitrate (NaNO3) is 85.00 g/mol (22.99 g/mol for sodium + 14.01 g/mol for nitrogen + 3 * 16.00 g/mol for oxygen).
Step 3: Determine the number of moles.
Calculate the number of moles for each reactant using the given masses and their respective molar masses:
Moles of CuCl2 = mass (g) / molar mass (g/mol) = 15 g / 134.45 g/mol
Moles of NaNO3 = mass (g) / molar mass (g/mol) = 20 g / 85.00 g/mol
Step 4: Use stoichiometry to compare moles.
From the balanced equation, we can see that the ratio between the number of moles of CuCl2 and NaNO3 is 2:2, so they have an equal stoichiometric ratio.
Step 5: Identify the limiting reactant.
Compare the moles of each reactant calculated in step 3. The reactant that produces fewer moles of product is the limiting reactant.
In this case, since the moles of CuCl2 and NaNO3 are equal, neither reactant limits the amount of product formed.
Therefore, the reaction will proceed until one of the reactants is completely consumed, and the amount of sodium chloride produced will be determined by the amount of the other reactant present in excess.