CHM 1045 CH 3 Homework Show all calculations, units, and conversion factors. For

Question

CHM 1045 CH 3 Homework Show all calculations, units, and conversion factors. For the reaction 2H2 + 102 2H20, there are 8.08 grams of H2 available. 1. Use molar mass (with two decimal places) to convert the grams of H2 into moles. NA to convert the moles of H2 into the number of molecules. Use significant figures. Show all units and conversion factors. (2 pts) Use the ratio of stoichiometric coefficients in the reaction above to determine how many moles of H20 can be produced from the H2 available. Then, use molar mass to convert the moles of H2O into grams. Use significant figures. Show all units and conversion factors. (2 pts) 2. Use the ratio of stoichiometric coefficients in the reaction above to determine how many moles of O2 will react with the H2 available. Then, use molar mass to convert the moles of O2 into grams. Use significant figures. Show all units and conversion factors. (2 pts) 3.

Explanation / Answer

Ans. #1. Moles of H2 = Mass / Molar mass

                                    = 8.08 g / (2.02 g/ mol)

                                    = 4.00 moles

No. of H2 molecules = Moles of H2 x Avogadro’s number

                                    = 4.00 moles x (6.022 x 1023 molecules / mole)

                                    = 2.41 x 1024 molecules

#2. According to the stoichiometry of balanced reaction, 2 mole H2 forms 2 mole H2O.

Hence,

Moles of H2O formed = (2 mol H2O / 2 mol H2) x 4.0 mol H2 = 4.00 mol H2O

            Mass of H2O formed = Moles of H2O formed x Molar mass

                                                = 4.00 mol x (18.02 g/ mol)

                                                = 72.08 g

#3. According to the stoichiometry of balanced reaction, 2 mole H2 reacts with 1 mole O2.

Hence,

Required moles of O2 = (1 mol O2/ 2 mol H2) x 4.0 mol H2 = 2.00 mol

            Required mass of O2 = 2.00 mol x (32.00 g/ mol) = 64.00 g

#4. According to the stoichiometry of balanced reaction, 2 mole H2 reacts with 1 mole O2.

We have,

            Available moles of H2 = 4.00 moles

            Available moles of O2 = 72.0 g/ (32.00 g/ mol) = 2.25 mol

            Theoretical requirement of moles of O2 = 2.00 moles (as calculated in #3)

Since the available moles of O2 is greater than its available moles while keeping moles of H2 constant, O2 is the reagent in excess. And, H2 is the limiting reactant.

# The yield of a reaction depends on the limiting reactant.

According to the stoichiometry of balanced reaction, 2 mole H2 forms 2 mole H2O.

We have, moles of limiting reactant H2 = 4.00 moles

Theoretical yield of H2O formed (moles) = 4.00 mol H2O        ; [see, #2]

            Theoretical yield of H2O formed (mass) = 4.00 mol x (18.02 g/ mol) = 72.08 g

#5. Actual yield of H2O = 64.0 g = 64.0 g / (18.02 g/ mol) = 3.55 mol

Now,

Theoretical yield (mass) = (Actual mass obtained / Theoretical mass produced) x 100

                                                = (64.0 g / 72.08 g) x 100

                                                = 88.79 %    

Theoretical yield (moles) = (Actual mol obtained / Theoretical mol produced) x 100

                                                = (3.55 mol / 4.00 g) x 100

                                                = 88.75 %    

The two values (% yield moles, and % yield mass) are nearly the same.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.