assuming this is a quanatiative reaction
the reaction is:
NaSO4(aq) + CaCl2(aq) --> CaSO4(s) + NaCl(aq)
0.50mol/L of sodium sulfate solution is meaured and put into 4 150ml beakers and 4 different volumes of 0.50mol/L of calcium chloride solution are prepared in a graduated cylinder. the caclium chloride will be added to the sodium solfate solution which is in the beaker and stirred.
find the limiting and excess reagents for each trial
trial | volume of 0.50mol/L of Na2SO4(aq) (mL) | Volume of 0.50mol/L CaCl2(aq) (mL) |
1) | 50 | 25
2) | 50 | 50
3) | 50 | 75
4) | 50 | 100
6 years ago
Answered By Leonardo F
Let's begin by writing the balanced chemical equation:
$Na_2SO_4_{\left(aq\right)}+CaCl_{2\left(aq\right)}\rightarrow CaSO_{4\left(s\right)}+2NaCl\left(aq\right)$Na2SO4(aq)+CaCl2(aq)→CaSO4(s)+2NaCl(aq)
We know that the porportion of sodium sulfate to calcium chloride is 1:1 (1 mol of sodium sulfate needs 1 mol of calcium chloride). This is the stoichiometric proportion between these two reagents.
Since the exercise informed us that the concentration of both reagents is the same (0.50 mol/L), the one with the biggest volume will have the most amount of moles, because:
$Concentration=\frac{moles}{volume}\rightarrow moles=concentration\times volume$Concentration=molesvolume →moles=concentration×volume
For the trial number 1, we have a bigger volume of sodium sulfate than calcium chloride. If we calculate the amount of moles of each reagent (remember the conversion 1 L = 1000 mL):
$moles\left(Na_2SO_4\right)=\left(\frac{0.50mol}{L}\right)\left(0.050L\right)=0.025mol$moles(Na2SO4)=(0.50molL )(0.050L)=0.025mol
$moles\left(CaCl_2\right)=\left(\frac{0.50mol}{L}\right)\left(0.025L\right)=0.013mol$moles(CaCl2)=(0.50molL )(0.025L)=0.013mol
Since the stoichiometric proportion of the reagents is 1:1, there is an excess of moles of sodium sulfate. Hence, sodium sulfate is the excess reagent, while calcium chloride is the limiting.
For trial number 2, we have the same volume of sodium sulfate and calcium chloride. Since their concentrations are the same (0.50 mol/L), we have the same amount of moles of the two reagents. Since the stoichiometric proportion of the reagents is 1:1, we don't have any excess or limiting reagents. We say that the reagents are in stoichiometric proportion.
For trial number 3, we have a bigger volume of calcium chloride than sodium sulfate. Since their concentrations are the same (0.50 mol/L), we don't have to calculate to know that we have more moles of calcium chloride than sodium sulfate. Then, calcium choride is in excess, while sodium sulfate is the limiting reagent.
For trial number 4, we have an even bigger volume of calcium chloride than sodium sulfate. Since their concentrations are the same (0.50 mol/L), we don't have to calculate to know that we have more moles of calcium chloride than sodium sulfate. Then, calcium choride is in excess, while sodium sulfate is the limiting reagent.