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# Chemistry Help!! Please?

I've got this experiment and here are the questions:

1. Record and calculate the following for the first half of the experiment:

(a) Mass of zinc used (g)

(b) Volume of 6M HCl used (mL)

(c) Molecular weight of zinc (g/mol)

(d) Moles of zinc reacted:

(e) Moles of hydrogen produced

(f) Molar volume of the ideal hydrogen gas at room temperature (Volume/moles), expressed as L/mol at X degrees C and a pressure of 1 atmosphere

1. For the second part of the experiment, everything was the same except that twice as much Zn was used. Record and calculate the following for this second reaction

(a) Mass of zinc used (g)

(b) Volume of 6M HCl used (mL)

(c) Molecular weight of zinc (g/mol)

(d) Moles of zinc reacted:

(e) Moles of hydrogen produced

(f) Molar volume of the ideal hydrogen gas at room temperature (Volume/moles), expressed as L/mol at X degrees C and a pressure of 1 atmosphere

1. Compare the molar volumes obtained in the two parts of the experiment.

What difference did using twice the amount of Zn in the second part?

Was all of the zinc used up in the reaction?

Which is the limiting reactant - zinc or HCl?

1. Compare the experimental value for the molar volume at 21C with the value listed in the Background section of the lab manual.

Calculate the experimental error according to

% error = | experimental molar volume - listed molar volume | / (listed molar volume) * 100

and some of the answers, but after the last one, I don't know how to answer??

A. .25 g

B. 10 ml

C. 65.39 g/ mol

D. 0.004 mol

E. After cooling down the gas volume was 102. 14 mL if 1 mol equals 22.4 L Then the moles are 0.005

F. 20.43 L/mol at 21 degrees C and 1 atm

A. 0.5 g Zn used

B. 10 mL

C. 65.39 g/mol

D. 0.008 mol

E. The gas volume for this one ended up being 194.28 and if 22.4 L is one mol then 194.28 mL = 0.009 mol

1. It about (not quite) doubled the amount of volume and moles produced. 