# 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 $C u C {l}_{2}$;

$0.672 L$ ${H}_{2}$;

$3.26 g C u \left(s\right)$;

$0.103 m o l H C l$ (or $3.74 g H C l$)

37.4%yieldCu(s)

#### Explanation:

Let's work with the copper(II) chloride solution problem. It says $154 m L$ of a $0.333 M$ solution of $C u C {l}_{2}$ reacted with $1.82 L$ ${H}_{2}$ 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 $C u$ (which it can, being just above $C u$ in the activity series). The chemical equation for this reaction is

$C u C {l}_{2} \left(a q\right) + {H}_{2} \left(g\right) \rightarrow C u \left(s\right) + 2 H C l \left(a q\right)$

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 $C u C {l}_{2}$ using the molarity equation:

$M = \text{mol solute"/"L soln}$ ; "mol solute" = (M)("L soln")

$m o l C u C {l}_{2} = \left(0.333 \frac{m o l}{\cancel{L}}\right) \left(0.154 \cancel{L}\right)$

= color(red)(0.0513molCuCl_2

Since one mole of gas occupies a volume of $22.4 L$ 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 $C u C {l}_{2}$ is less than that of ${H}_{2}$, so the limiting reagent is thus $C u C {l}_{2}$. This means all of it will be used up, and therefore no $C u C {l}_{2}$ will remain at the end of the reaction.

The amount of ${H}_{2}$ remaining at the end of the reaction is

$\textcolor{b l u e}{0.0813 m o l} - \textcolor{red}{0.0513 m o l} = \textcolor{g r e e n}{0.0300 m o l {H}_{2}} = \textcolor{g r e e n}{0.672 L {H}_{2}}$

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

The amount of $C u \left(s\right)$ 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 $H C l \left(a q\right)$ remaining after the reaction is calculated the same way:

$\textcolor{red}{0.0513 \cancel{m o l C u C {l}_{2}}} \left(\frac{2 \cancel{m o l H C l}}{1 \cancel{m o l C u C {l}_{2}}}\right) \left(\frac{36.46 g H C l}{1 \cancel{m o l H C l}}\right) = \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{3.74 g H C l}$

Calculating the mass of $H C l$ remaining isn't that conventional, as it is dissolved in solution. If the water were evaporated, and the $H C l$ gas was separated from the evaporated ${H}_{2} O$, the mass of the pure $H C l$ gas would be $\textcolor{p u r p \le}{3.74 g}$. You could just report the remaining quantities in moles, which is just $\textcolor{red}{0.0513 m o l} \times 2 = \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{0.103 m o l H C l}$.

Lastly, let's calculate the percent yield of the $C u \left(s\right)$, 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 $1.22 g C u \left(s\right)$, the percent yield of $C u \left(s\right)$ 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).