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a Secure | https:ffsession.masteringchemistry.com/myctitem ViewassignmentProbiemID 88374653 Exercise 4.34 Enhanced with Feedback Exercise 434- Enhanced with Feedback Part A For each of the reactions, calculate the mass (in grams) of the product formed when 15.36 g of the underlined reactant completely reacts. Assume that there is more than enough of the 4) Section 4.2 While completing this problem Incorrect: Try Again: 3 attempts remaining Part B 2 Kis)Bral)KBrs) Part C :19 PM Search the web and Windows

Explanation / Answer

Part A

Given reaction is

2 K (s) + Cl2 (g) ---> 2 KCl (s)

2 mole 1 mole 2 mole

Given

Mass of Cl2 = 15.36 g

Molar mass of Cl2 = 70.9 g/mol

No. of moles of Cl2 = Mass / Molar mass = 15.36 g / 70.9 g/mol = 0.217 moles of Cl2

According to reaction stoichiometry 1 mole of Cl2 will produce 2 moles of KCl

0.217 moles of reacting with excess K will produce 0.433 moles of KCl

Molar mass of KCl = 74.55 g/mol

Mass of product formed = No. of moles * Molar mass = 0.433 moles * 74.55 moles = 32.3 g Answer

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Part B

Given reaction is

2 K (s) + Br2 (g) ---> 2 KBr (s)

2 mole 1 mole 2 mole

Given

Mass of Br2 = 15.36 g

Molar mass of Br2 = 159.81 g/mol

No. of moles of Br2 = Mass / Molar mass = 15.36 g / 159.81 g/mol = 0.0961 moles of Br2

According to reaction stoichiometry 1 mole of Br2 will produce 2 moles of KBr

0.0961 moles of reacting with excess K will produce 0.1922 moles of KBr

Molar mass of KBr = 119 g/mol

Mass of product formed = No. of moles * Molar mass = 0.1922 moles * 119 moles = 22.875 g Answer

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