A 0.495 g sample of cobalt, Co, is added to 183 ml of 0.100 M H2SO4 to produce c

Question

A 0.495 g sample of cobalt, Co, is added to 183 ml of 0.100 M H2SO4 to produce cobalt (II) sulfate and hydrogen gas. Co(s) + H,so,(aq) Coso, (aq) + H,(g) What is the limiting reactant? H2SO4(aq) Co(s) A volume of 103 ml of hydrogen is collected over water. The water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 101.325 kPa, the temperature is 25 °C, and the vapor pressure of water at 25 °C is 3.167 kPa. What is the percent yield of hydrogen gas for this reaction? Number

Explanation / Answer

1)

mass of Co = 0.495 g

moles of Co = 0.495 / 58.93 = 8.4 x 10^-3 mol

moles of H2SO4 = 183 x 0.1 / 1000 = 0.0183 mol

Co    + H2SO4    -------------> CoSO4   +   H2

1             1

8.4 x 10^-3       0.0183

here limiting reagent is Co

2)

pressure = 101.325 - 3.167 = 98.158 Kpa

               = 0.9687 atm

temperature = 298 K

moles of H2 = moles of Co = 8.4 x 10^-3

P V = n R T

0.9687 x V = 8.4 x 10^-3 x 0.0821 x 298

V = 0.212 L

volume of H2 = 212 mL

actual yield of H2 = 103 mL

% yield = 103 / 212 ) x 100

% yield = 48.6 %

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