If anyone could help me out that would be amazing,with hemi

Question

What is the greatest amount of sodium chloride in moles that can be made with 13 grams copper II chloride and 15 grams sodium nitrate? Which reactant is limiting? How many moles of the excess reactant will be left after the reaction?

CuCl₂  + 2 NaNO₃  --->Cu(NO₃)₂ + 2 NaCI

Answer 1

Hi Emily! There's a lot of steps to solving these types of problems and they can sometimes be tricky, so I will break down all of these steps for you as clearly as possible so you can understand how to solve limiting reactant problems.

Step 1: Make sure the reaction equation is balanced (there should be equal amounts of each atom present on the reactants and products side of the equation). The equation you provided appears to be balanced so we can continue onto the next step.

Step 2: Determine the limiting reactant. The limiting reactant is going to be the substance on the reactants side that will be 100% consumed by the time the reaction is complete. Before we determine which reactant is limiting, its easier if we convert the masses of each reactant into moles, which can be done using the molar mass of each reactant. You also need to understand how to write conversion factors. The general formula for a conversion factor is :

Given unit x (Desired Unit / Given Unit) = Desired Unit

For CuCl₂: 13 grams (1 mole CuCl₂ / 134.45 grams) = 0.0967 moles of CuCl₂ available for this reaction

For NaNO₃: 15 grams (1 mole NaNO₃ / 84.99 grams) = 0.1765 moles of NaNO₃ are available for this rxn.

In order to determine which of these reactants is limiting, we first have to figure out how many moles of one reactant it would take to completely consume the other reactant. For this, you need the stoichiometric coefficients from the balanced reaction equation, which is why you should always make sure your equation is properly balanced before doing any calculations. Based on the equation, it takes 1 mole of copper (II) chloride and 2 moles of sodium nitrate to form 1 mole of copper nitrate and 2 moles of sodium chloride.

Now let's assume that the copper (II) chloride is completely consumed:

0.0967 moles CuCl₂ ( 2 moles NaNO₃ / 1 mole CuCl₂ )

= 0.1934 moles of NaNO₃ would be required to completely consume your 13 grams of CuCl₂

Now let's assume that your NaNO₃ is completely consumed:

0.1765 moles NaNO₃ ( 1 mole CuCl₂ / 2 moles NaNO₃ )

= 0.08825 moles of CuCl₂ would be required to completely consume your 15 grams of NaNO₃

Now look back at your available molar quantities for each reactant. You only have 0.0967 moles of CuCl₂, and 0.1765 moles of NaNO₃ available for your reaction. If the CuCl₂ completely reacted, it would take 0.1934 moles of NaNO₃ (which you do not have). However, you do have enough moles of CuCl₂ available to consume the 0.1765 moles of NaNO₃ completely (because only 0.08825 moles are needed and you have 0.0967 moles). Since you have the ability to completely consume all of your NaNO₃, the NaNO₃ will be your limiting reactant, and you will have an excess of the CuCl₂. Now that we know the limiting reactant, we can do the last two calculations.

Step 3: Calculate the maximum amount of NaCl that can be produced. Since the limiting reactant determines how much product is formed, all we need to do is convert the mass of NaNO₃ into moles, then use the stoichiometric coefficients from the balanced equation to relate the moles of NaNO₃ being consumed to the moles of NaCl being produced, and then we can calculate the maximum theoretical mass of NaCl that is produced.

15 grams NaNO₃ (1 mole NaNO₃/84.99 grams)(2 moles NaCl/2 moles NaNO₃)(58.44 grams/1 mole NaCl)

= 10.3142 grams of NaCl should be produced if the NaNO₃ reacts completely

Step 4: Determine the moles of excess reactant. This can be done by subtracting the moles of the excess reagent that were consumed from the total moles of the excess reagent that you had before the reaction started. We started off with 0.0967 moles of CuCl₂ and ended up consuming 0.08825 moles of CuCl₂ (I got this number from step 2 when I did the limiting reactant calculations).

0.0967 moles - 0.0883 moles = 0.0084 moles of CuCl₂ remain (aka your excess amount)

I hope this helps! Please be sure to leave feedback or comment if you have any questions or need any clarification!