Zn(s) + HCL(aq) ----> Zn 2+(aq) +Cl-(aq) + H2(g) + heat. Balance the equation an
ID: 979279 • Letter: Z
Question
Zn(s) + HCL(aq) ----> Zn 2+(aq) +Cl-(aq) + H2(g) + heat.
Balance the equation and:
a)calculate the number of moles of hydrogen gas that can be produced from 1.0g of zinc (6.38g/mol) pellets, assuming the reaction completes.
b)Calculate the final temperature of the solution of 10.0 mL of 1.0 M aqueous HCl at an initial temperature of 20.7 degrees Celsius was used to consume the zinc pellets. Use deltaHrxn= -80.7 kJ/mol and assume that the hydrogen gas doesn't affect the temperature.
c) Calculate the volume of hydrogen gas collected at 1 atm pressure and the temperature from part b and ignore the contribution of water vapor.
Explanation / Answer
Solution :-
Balanced reaction equation is as follow
Zn(s) + 2HCL(aq) ----> Zn 2+(aq) +2 Cl-(aq) + H2(g) + heat.
a)calculate the number of moles of hydrogen gas that can be produced from 1.0g of zinc (6.38g/mol) pellets, assuming the reaction completes.
Solution :- using the mole ratio of the Zn and H2 we can find the moles of the H2 produced (1.0 g Zn * 1 mol / 65.39 g )*(1 mol H2/ 1 mol Zn) = 0.015293 mol H2
b)Calculate the final temperature of the solution of 10.0 mL of 1.0 M aqueous HCl at an initial temperature of 20.7 degrees Celsius was used to consume the zinc pellets. Use deltaHrxn= -80.7 kJ/mol and assume that the hydrogen gas doesn't affect the temperature.
Solution :- T1 = 20.7 C
Volume of solution = 10 ml
Moles of HCl = 1.0 mol per L * 0.010 L = 0.010 mol HCl
0.01mol HCl * 1 mol H2 / 2 mol HCl = 0.005 mol H2
q = 80.7 kJ * 0.005 mol / 1 mol = 0.4035 kJ
0.4035 kJ * 1000 J / 1 kJ= 403.5 J
Now lets calculate the change in temperature
Assume density of solution is 1 g/ml then mass of solution is 10 g and specific heat of solution = 4.184 J per g C
q= m*c*delta T
q/m*c = delta T
403.5 J / 10 g * 4.184 J per g C = delta T
9.64 C= delta T
So the final temperature is 20.7 C + 9.64 C = 30.34 C
c) Calculate the volume of hydrogen gas collected at 1 atm pressure and the temperature from part b and ignore the contribution of water vapor.
Solution :- p= 1 atm
V= ?
n=0.005 mol H2
T= 30.34 C +273 = 303.34 K
PV= nRT
V= nRT / P
= 0.005 mol * 0.08206 L atm per mol K *303.34 K / 1 atm
= 0.1245 L
0.1245 L * 1000 ml / 1 L = 124.5 ml
So the volume of the H2 gas formed = 124.5 ml
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