Predicting Products & Reaction Types Lab Report

Purpose: To predict products of five reaction types and perform them

Hypothesis
Station 1: Mg+ O2 -> 2MgO
Station 2: NaHCO3 -> Na2CO3 + H2O
Station 3: Pb(NO3)2 + KI -> PbI2 + KNO3
Station 4: CH3OH + O2 -> CO2+ H2O
Station 5: CuCl2 + Al -> AlCl2 + Cu

Data

Analysis Questions

Station 1

• Write the balance equation for this reaction, including the state of matter for each reagent.

2Mg(s) + O₂(g)->2MgO(s) + energy

• Which type of reaction (of the 5 major types) is this? Composition reaction
• Calculate the percent yield of the product.

(0.038g Mg)/1*(1mol Mg)/(24.305g Mg)*(2mol MgO)/(2mol Mg)*(40.304g MgO)/(1mol MgO)=0.063g MgO

actual/theoretical*100=0.021/0.063*100=33%

• Provide an explanation for the percent yield.

The percent yield is low because the reaction was not complete; I still had some magnesium left and I probably lost mass while measuring the mass of the magnesium after the reaction.

Station 2

• Write the balance equation for this reaction, including the state of matter for each reagent.

2NaHCO3(s)->Na2CO3(s) + H2O(g) + CO2(g)

• Which type of reaction (of the 5 major types) is this? Decomposition reaction
• Calculate the mass of each product.

(1.63g NaHCO3)/1*(1mol NaHCO3)/(84.006g NaHCO3)*(1mol Na2CO3)/(2mol NaHCO3)*(105.998g Na2CO3)/(1mol Na2CO3)=1.028g Na2CO3

(1.63g NaHCO3)/1*(1mol NaHCO3)/(84.006g NaHCO3)*(1mol H2O)/(2mol NaHCO3)*(33.006g H2O)/(1mol H2O)=0.320g H2O

(1.63g NaHCO3)/1*(1mol NaHCO3)/(84.006g NaHCO3)*(1mol CO2)/(2mol NaHCO3)*(44.009g CO2)/(1mol CO2)=0.427g CO2

• Compare the theoretical masses of your products to the actual mass of the reactant and discuss the comparison. Which natural law applies? They are very similar with a difference of 0.1g and this indicates the Law of Conservation of Mass.

Station 3

• Write the balance equation for this reaction, including the state of matter for each reagent.

Pb(NO3)2(aq) + 2KI(aq)->2KNO3(aq) + PbI2(s)

• Which type of reaction (of the 5 major types) is this? Double-replacement reaction
• Calculate the number of moles used of each reactant (using the concentrations).

(2mL Pb(NO3)2)/1*(1L Pb(NO3)2)/(1000mL Pb(NO3)2)*(0.1 mol Pb(NO3)2)/(1L Pb(NO3)2)=2×〖10〗^(-4) mol Pb(NO3)2

(0.5mL KI)/1*(1L KI)/(1000mL KI)*(0.1mol KI)/(1L KI)=5×〖10〗^(-5) mol KI

• How many moles of each product was made?

(2×〖10〗^(-4) mol Pb(NO3)2)/1*(2mol KNO3)/(1mol Pb(NO3)2)=4×〖10〗^(-4) mol KNO3

(2×〖10〗^(-4) mol Pb(NO3)2)/1*(1mol PbI2)/(1mol Pb(NO3)2)=2×〖10〗^(-4) mol PbI2

• Which product is the precipitate? Which physical property facilitated its formation?
  Lead Iodide is the precipitate. Its solubility facilitated its formation.

Station 4

• Write the balance equation for this reaction, including the state of matter for each reagent.

2CH3OH(l) + 3O2(g)-> 2CO2(g) + 4H2O(l)

• Which type of reaction (of the 5 major types) is this? Combustion reaction
  What is the percent composition of methanol?

C=2×12.011=24.022/64.084=.375×100=37.5%
H=8×1.008=8.064/64.084=.126×100=12.6%
O=2×15.999=31.998/64.084=.499×100=49.9%

Station 5

• Write the balance equation for this reaction, including the state of matter for each reagent.

3CuCl2(aq) + 2Al(s)->2Cu(s) + 2AlCl3(aq)

• Which type of reaction (of the 5 major types) is this? Single-replacement reaction
• Calculate the number of moles of copper (II) chloride used in this experiment.

(10mL CuCl2)/1*(1L CuCl2)/(1000mL CuCl2)*(1.0mol CuCl2)/(1L CuCl2)=0.01mol CuCl2

• Which reactant is the limiting reactant? Support this answer mathematically for credit.

(0.01mol CuCl2)/1*(2mol Cu)/(1mol CuCl2)=0.2mol Cu

(1.63g Al)/1*(1mol Al)/(26.982g Al)*(2mol Cu)/(2mol Al)=0.06mol Cu

Aluminum is the limiting reactant.

William Anderson (Schoolworkhelper Editorial Team)

William completed his Bachelor of Science and Master of Arts in 2013. He current serves as a lecturer, tutor and freelance writer. In his spare time, he enjoys reading, walking his dog and parasailing. Article last reviewed: 2022 | St. Rosemary Institution © 2010-2024 | Creative Commons 4.0

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