A 0.645 g sample of cobalt, Co, is added to 239 mL of 0.100 M H, SO, to produce cobalt(II) sulfate and hydrogen gas.Co(s) + H,SO,(aq)→ CoSO,(aq) + H,(g)What is the limiting reactant?Co(s)OH, SO, (aq)A volume of 153 mL of hydrogen is collected over water. The water level in the collecting vessel is the same as the outsidelevel. Atmospheric pressure is 101.325 kPa, the temperature is 25 °C, and the vapor pressure of water at 25 °C is3.167 kPa.What is the percent yield of hydrogen gas for this reaction?percent yield:%

Answered: A 0.645 g sample of cobalt, Co, is…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Question

A 0.645 g sample of cobalt, Co, is added to 239 mL of 0.100 M H, SO, to produce cobalt(II) sulfate and hydrogen gas.
Co(s) + H,SO,(aq)
→ CoSO,(aq) + H,(g)
What is the limiting reactant?
Co(s)
O
H, SO, (aq)
A volume of 153 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?
percent yield:
%

Expert Solution

Step 1

Answer - Here I have explained the step by step solution . Hope this will really gonna help you !!!!!

So lets calculate the no of moles of cobalt (Co) = mass of cobalt / molar mass of cobalt

moles of cobalt = 0.645 g / 58.93 g-mol^-1 = 0.0109 mol

moles of H₂SO₄ = 0.100 M * 239 mL * 0.001 L/mL = 0.0239 mol

Now according to the balanced reaction Co and H2SO4 will react in 1:1 mole ratio . But Co is in lesser (mol) amount . So the limiting reagent will be = Co (Cobalt)

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Sample Preparation in Analytical Chemistry

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions