Problem 2. Consider the following reaction, which proceeds to completion. 1 CrC0

Question

Problem 2. Consider the following reaction, which proceeds to completion. 1 CrC03(aq) + 2 HBrO)(aq) 1 Cr(Bro,)2 (aq) + 1 COz(g) + 1 H20(1) 28.08 g of CrCO reacts with 481 mL of 0.969 M HBrOg. The temperature is 34.9 °C and total pressure is 18.2 psi? The vapor pressure of water at this temperature is 42.35 torr. a. Write the net ionic equation. b. What ion is limiting? c. How many moles of the excess ion remains? d. How many moles of each spectator ion remains? e. What is the partial pressure of CO2, in atm? f What volume of dry CO2, in cm, is produced?

Explanation / Answer

Part a

Molecular ionic equation

Cr2+ + CO3 2- + 2H+ + 2BrO3- = Cr2+ + 2BrO3- + CO2 + H2O

Net ionic equation

CO3 2- + 2H+ = CO2 + H2O

Part b

Moles of CrCO3 = mass/molecular weight

= 28.08/112 = 0.2507 mol

Moles of HBrO3 = molarity x volume

= 0.969 mol/L x 0.481 L

= 0.466 mol

From the stoichiometry of the reaction

1 mol CrCO3 reacts with = 2 mol of HBrO3

0.2507 mol CrCO3 reacts with = 2*0.2507 = 0.5014 mol of HBrO3

But we have only 0.466 mol of HBrO3 than required 0.5014 mol of HBrO3

Limiting reactant = HBrO3

Excess reactant = CrCO3

Limiting ion = H+

Excess ion = CO3 -2

Part C

Moles of excess ion remain (CO3 2-) = Initial - reacted

= 0.2507 - (0.466/2) = 0.0177 mol

Part d

Spectator ions = Cr2+ & BrO3-

Moles of Cr2+ remain = Moles of excess ion remain (CO3 2-)

= 0.0177 mol

Moles of BrO3- remain = 0

Part e

Pressure of dry CO2 gas = Barometric pressure - Vapor pressure of water

= [18.2 psi x 1 atm/14.7 psi] - [ 42.35 torr x 1 atm/760 Torr]

= 1.238 - 0.0557 = 1.1823 atm

Part f

From the ideal gas equation

V = nRT/P

= [(0.466/2) mol x 0.0821 L-atm/mol-K x (34.9 + 273.15)K]/[1.1823 atm]

= 4.984 L x 1000 cm3/L

= 4984 cm3

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