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?

Question

Mateo Aguilar · Accepted Answer

No #CuCl_2#;

#0.672L# #H_2#;

#3.26gCu(s)#;

#0.103molHCl# (or #3.74gHCl#)

#37.4%yieldCu(s)# 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)# The task at hand involves determining the limiting reactant whenever two or more are given in quantities. 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 the reaction uses the same amounts of both reactants. 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 percent yield formula 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 particularly high, but it serves as an illustration:) I hope this was helpful, and I apologize if it seemed overwhelming—most of it is really simple.

Mateo Aguilar · Answer

undefined To determine the amount of each substance present at the end of the reaction and the percent yield, you need to know the balanced chemical equation and the experimental yield.

Without the balanced chemical equation and additional information about the reaction and the substances involved, it's not possible to provide a specific answer. Could you please provide more details or the balanced chemical equation for the reaction in question?