# CHEM125 – CHAPTER 04 – REGULAR (LIMITING REAGENT)

## CHAPTER 4 REGULAR (Limiting Reagent) HOMEWORK

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### Question 1

The reaction of 9.75 g of Cl2 with 2.20 g of P4 produces 4.35 g of PCl5. What is the percent yield?
Hint given in feedback ### Explanation

Balanced Equation P4 + 10Cl2 —-> 4PCl5 moles P4 = mass / molar mass = 2.20 g / 123.88g/mol = 0.01776 mol 1 mole P4 reacts to produce 4 moles PCl5 So moles PCl5 possible = 4 x moles P4 = 0.07104moles PCl5 possible moles Cl2 = 9.75 g / 70.9 g/mol = 0.1375 mol 10 mol Cl2 react to form 4 moles PCl5 Therefore moles PCl5 possible = 4/10 x moles Cl2 = .055 mol PCl5 possible (select the one with the least amount of output, in this case Cl2 is the limiting reagent) .055 mol PCl5 * ( 208g PCl5/ mol) = 11.45 g (theoretical) actual = 4.35g PCl5 (given) 4.35g / 11.45g = 0.38 0.38 x 100= 38.0 % (don’t include percent sign for moodle)

### Question 2

A chemical reaction produces 3.5 g of the product. If the theoretical yield is 5.9 g what is the % yield? ### Explanation 3.5g / 5.9g = .593 .593 x 100 = 59.3 %

### Question 3

Calculate the mass (in g) of O2 consumed in the complete combustion of 27.4 g sample of C4H8O. (Hint, write the balanced equation.) ### Explanation

Balanced equations 2 C4H8O +11 O2 = 8 CO2 + 8 H2O moles C4H8O = 27.4 g / 72.108 g/mol=0.379 the ratio between C4H8O and O2 is 2 : 11 moles O2 needed = 0.379 x 11 /2 = 2.089 mass O2 = 2.089 mol x 32 g/mol=66.9 g ( 3 significant figures )

### Question 4

Carbon disulfide (CS2) reacts with excess chlorine (Cl2) to produce carbon tetrachloride (CCl4) and disulfur dichloride (S2Cl2). If 61.1 g of CS2 yields 79.6 g of CCl4, what is the percent yield? (Hint, you must first write the balanced equation.) ### Explanation

Balanced Equation: CS2 + 3 Cl2 –> CCl4 + S2Cl2 Theoretical Yield: 61.1g CS2 * (1 mol CS2/ 76.139 g CS2) * (1 mol CCl4/ 1mol CS2) * (153.82 g CCl4/ 1 mol CCl4) = 123.4 g CCl4 Actual Yield : 79.6g CS2 Percent of Yield: 79.6/123.4g = .645 x 100 = 64.5 %