Molar volume at room conditions 7,8,9 Mass of Zinc used .20g .20g/ 65.38=.0031 F

Question

Molar volume at room conditions 7,8,9

Mass of Zinc used .20g .20g/ 65.38=.0031 Final temperature (room temperature) 294.6K Final pressure 1.53atm Calculations - for each of following do the necessary calculation to obtain the item requested. Place your calculated result to the right of the item. Be sure to include the proper units. In the space below the item describe how you calculated your answer. (see end of experiment 3 assignment for suggestions on how to describe a calculation. Moles of Zn reacted .20g/65.38=.0031 5. Moles of Zn calculated above is also moles H_2 produced. Explain why this is true. the ratio of is 1:1 so .0031 mol of H2 are produced from the reaction Volume of gas collected at room conditions. (.0031 Times .08206 times 294.6) / 1 = .0749mL Molar volume at room conditions. Molar volume at STP. Using the expected or accepted Molar Volume of H_2 at STP (check background in lab manual), compare your results to the accepted value by calculating the percent error. Percent error = ((your value - accepted value)/accepted value) Times 100

Explanation / Answer

Solution :-

7) Moles of H2 produced = 0.031 mol

Volume of the H2 at room conditions = V= nRT/ P

                                                       = 0.0031 mol * 0.08206 L atm per mol K * 294.6 K / 1.53 atm

                                                       = 0.04898 L

8) Molar volume at room temperature = 0.04898 mol / 0.0031 mol = 15.8 L

At STP temeorature is 273 K and pressure is 1 atm

So V= nRT/ P

      = 0.0031 mol * 0.08206 L atm per mol K*273 K / 1 atm

      = 0.06945 L

Molar volume = 0.06945 L / 0.0031 mol = 22.4 L

Q9 lets calculate the error

% error = (22.4 L – 15.8 L / 22.4 L)*100%

             = 29.46 %

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