how much of each substance will be present at the end of the reaction? If 1.22g of the solid formed, what is the percent yield of this reaction?

1 Answer
May 21, 2017

No CuCl_2;

0.672L H_2;

3.26gCu(s);

0.103molHCl (or 3.74gHCl)

37.4%yieldCu(s)

Explanation:

Let's work with the copper(II) chloride solution problem. It says 154mL of a 0.333M solution of CuCl_2 reacted with 1.82L H_2 at standard temperature and pressure. Ideally, for single-replacement reactions such as this, you'll need to look at a table of standard reduction potentials (or an activity series) to see if H_2 will be able to displace Cu (which it can, being just above Cu in the activity series). The chemical equation for this reaction is

CuCl_2(aq) + H_2(g) rarr Cu(s) + 2HCl(aq)

Whenever you're given quantities of two (or more) reactants, you'll be finding which one is limiting.

To find the limiting reactant, you'll need to find the number of moles of each reaction is present, and then divide the number by the coefficient in front of it. Since the coefficients for both reactants are 1, we don't need to divide any further, but if the coefficients were different, we would need to divide them and, for either case, the reactant which (after dividing) has the lowest number of moles is limiting.

Let's calculate the moles of CuCl_2 using the molarity equation:

M ="mol solute"/"L soln" ; "mol solute" = (M)("L soln")

molCuCl_2 = (0.333(mol)/(cancelL))(0.154cancelL)

= color(red)(0.0513molCuCl_2

Since one mole of gas occupies a volume of 22.4L at stp, the number of moles of H_2 is

1.82cancel(LH_2)((1molH_2)/(22.4cancel(LH_2))) = color(blue)(0.0813molH_2

We can see that the quantity of CuCl_2 is less than that of H_2, so the limiting reagent is thus CuCl_2. This means all of it will be used up, and therefore no CuCl_2 will remain at the end of the reaction.

The amount of H_2 remaining at the end of the reaction is

color(blue)(0.0813mol) - color(red)(0.0513mol) = color(green)(0.0300 molH_2) = color(green)(0.672LH_2)

because equal amounts of both reactants are used in the reaction.

The amount of Cu(s) remaining can be found by using the color(red)(0.0513molCuCl_2 calculated and the stoichiometric relationships in the equation. Always use the moles of the limiting reactant to calculate the quantities of other substances.

color(red)(0.0513cancel(molCuCl_2))((1cancel(molCu))/(1cancel(molCuCl_2)))((63.55gCu)/(1cancel(molCu))) = color(purple)(3.26gCu

There will thus be color(purple)(3.26gCu remaining after the reaction goes to completion.

The amount of HCl(aq) remaining after the reaction is calculated the same way:

color(red)(0.0513cancel(molCuCl_2))((2cancel(molHCl))/(1cancel(molCuCl_2)))((36.46gHCl)/(1cancel(molHCl))) = color(darkorange)(3.74gHCl)

Calculating the mass of HCl remaining isn't that conventional, as it is dissolved in solution. If the water were evaporated, and the HCl gas was separated from the evaporated H_2O, the mass of the pure HCl gas would be color(purple)(3.74g). You could just report the remaining quantities in moles, which is just color(red)(0.0513mol) xx 2 = color(darkorange)(0.103molHCl).

Lastly, let's calculate the percent yield of the Cu(s), which (like all of this, even though I might make it seem that way) is not too difficult.

The equation for percent yield is

%yield = a/t xx 100%

where a represents the actual yield (the amount obtained) and
t represents the theoretical yield (the amount calculated) of the substance.

Since we obtained 1.22gCu(s), the percent yield of Cu(s) is

(1.22cancel(gCu(s)))/(3.26cancel(gCu(s))) xx 100%

= color(violet)(37.4%yieldCu(s)

Which isn't very high, but it's just an example problem :)

I hope this was useful, and sorry if it was overwhelming (it really is straightforward for the most part).