Question

1) How many liters of carbon dioxide gas will be produced at 32°C and 747 mmHg if 1.595 kg of calcium carbonate decomposes? (calcium oxide is the other product)
2) If 175.0mL of a 2.12M potassium chloride solution is mixed with 95.0mL of a 5.97M lead (II) nitrate solution, how many grams of precipitate will be produced in this reaction?

Answer 1

Answer:

1. $V_{CO_2}=405.85L$

2. $m_{PbCl_2}=51.6gPbCl_2$

Explanation:

Hello,

1) In this case, the reaction is:

$CaCO_3\rightarrow CaO+CO_2$

Thus, since 1.595 kg of calcium carbonate, whose molar mass is 100 g/mol, is decomposed, the following moles of carbon dioxide are produced:

$n_{CO_2}=1.595kgCaCO_3*\frac{1000gCaCO_3}{1kgCaCO_3}*\frac{1molCaCO_3}{100gCaCO_3} *\frac{1molCO_2}{1molCaCO_3} \\\\n_{CO_2}=15.95molCO_2$

Then, by using the ideal gas equation we can compute the volume of carbon dioxide:

$V_{CO_2}=\frac{n_{CO_2}RT}{P}=\frac{15.95molCO_2*0.082\frac{atm*L}{mol*K}*(32+273)K}{747mmHg*\frac{1atm}{760mmHg} } \\\\V_{CO_2}=405.85L$

2) In this case, the reaction is:

$2KCl(aq)+Pb(NO_3)_2(aq)\rightarrow PbCl_2(s)+2KNO_3(aq)$

Thus, lead (II) chloride is the precipitate. In such a way, we first identify the limiting reactant by firstly computing the moles of potassium chloride and lead (II) nitrate with the volumes and the molarities:

$n_{KCl}=2.12mol/L*0.175L=0.371molKCl\\n_{Pb(NO_3)_2}=5.97mol/L*0.095L=0.567molPb(NO_3)_2$

Next, we compute the moles of potassium chloride consumed by 0.567 moles of lead (II) nitrate by using their 2:1 molar ratio:

$n_{KCl}^{consumed\ by\ Pb(NO_3)_2}=0.567molPb(NO_3)_2*\frac{2molKCl}{1molPb(NO_3)_2}=1.134molKCl$

Therefore, as just 0.371 moles of potassium chloride are available we say it is the limiting reactant, for that reason, the grams of lead (II) chloride, whose molar mass is 278.1 g/mol, precipitate finally result:

$m_{PbCl_2}=0.371molKCl*\frac{1molPbCl_2}{2molKCl}*\frac{278.1gPbCl_2}{1molPbCl_2} \\\\m_{PbCl_2}=51.6gPbCl_2$

Best regards.