When 0.0150 mol of a metal was reacted with 30.0 mL of 4.00 M HC1, 1150 mL of wa

Question

When 0.0150 mol of a metal was reacted with 30.0 mL of 4.00 M HC1, 1150 mL of water were displaced at a barometric pressure of 755.0 torr. The temperature of both the H_2 and H_2O were 26.0 degree C. The excess HC1 left from the reaction was then titrated with NaOH. 25.00 mL of 1.20 M NaOH were required to neutralize the excess HC1. Given the above information, answer the following questions: a) How many mol of HC1 were initially used in the reaction? b) How many mol of HC1 were in excess? c) How many mol of HCl actually reacted with the metal? d) What is the partial pressure of H_2 (in torr) that was produced? (You will need to use Table A in Appendix VI). e) How many mol of H_2 were produced? f) Fill in the blanks to the following equation: 1 M + ___ HCI rightarrow ___ MCI ___ + ____ H_2

Explanation / Answer

a)   Amount of HCl was used = 30 mL of 4.00 M = (30 x 10-3 L ) x 4 mol/L = 0.120 mol
b)   Volume of solution, V1 = 1150 mL
Molarity of HCl in the solution = M1
Volume of NaOH, V2 = 25 mL
Molarity of NaOH, M2 = 1.2 M

M1V1 = M2V2
M1 = M2V2/V1 = (25 x 1.2) /1150 = 0.026 M
Moles of HCl = 1.150 L x 0.026 M = 0.03 mol
0.03 mol of HCl in excess.
c) Moles of HCl reacted with the metal = 0.120 – 0.03 = 0.09 mol
f) 0.0150 mol of metal react with 0.09 mol of HCl
1 mol of metal react with = 0.09/0.015 = 6 mol HCl
1M + 6 HCl ------- 6 MCl + 3 H2
e) mol of H2 produced = 3 x 0.015 = 0.045 mol
d) number of H2 = 0.045 mol
Volume of water = 1150 mL = 1150 g d = 1 g/mL
number of moles of water =1150/18 = 63.9 mol
mole fraction of H2 = 0.045/63.945 = 0.0007
partial pressure of H2 = mole fraction X total pressure = 0.0007 x 755 = 0.53 torr

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