PROBLEM 5 The Joker has decided to throw a party to welcome Batman\'s new sideki
ID: 553729 • Letter: P
Question
PROBLEM 5 The Joker has decided to throw a party to welcome Batman's new sidekick, Diclk Grayson. In an attempt to end the Joker's destructive antics, Commissioner Gordon has been monitoring all purchases of dangerous materials. Therefore, Joker must improvise using the materials he already has Using leftover HCl from the vat he used to kill the last Robin, and some ordinary bleach, Joker plans to trap Batman in the party room with balloons filled with deadly chlorine gas. Joker will run his reaction in an airtight reaction vessel, bubbling the gas over water, connected to the balloons directly (nearly identical setup as used in lab 3) If Joker wants twelve 2.5 L balloons filled with chlorine gas, how many mL of bleach does he need to use? Assume that HCl is in high excess, the atmospheric pressure is 1.01 atm, and it is 80 °F in the Joker's hideout. NaOCI(aq) + 2 HCl (aq) Cl2 (aq) + H20() + NaCl(aq)Explanation / Answer
NaOCl(aq) + 2HCl(aq) -----> Cl2(aq) + H2O(l) + NaCl (aq)
In this reaction one mole of NaOCl reacts with HCl and produces one mole of Cl2 gas.
Since we are using HCl in excess, no moles of NaOCl = no. moles of Cl2 gas
From ideal gas PV = nRT
no. moles n = PV/RT
Pressure = 1.01 atm ; Volume V = 2.5 x 12 = 30L
R = 0.08205 L atm / mol·K ; T = T(°K) = (80(°F) + 459.67)× 5/9 = 300° K
no. moles n = PV/RT
= (1.01 atm x 30L)/( 0.08205 L atm / mol·°K ) (300°K)
= 1.23 moles
No. of moles of Cl2 gas = 1.23 moles
No. moles of NaOCl = 1.23 moles.
Mass of NaOCl = no. moles x mol.wt
= 1.23 x 74.44 g/mol
= 91.63 g
Therefore, we need 91.63 g of NaOCL. we need to find how many mL required. The density of NaOCl= 1.11 g/mL
Means; 1mL of NaOCl weigh ......1.11 g
Therefore, we need to find 91.63 g of NaOCl how much it weighs;
Volume of NaOCl required = 91.63 g x 1 mL / 1.11 g
= 82.55 mL
Hence, the volume of NaOCl required = 82.55 mL
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.