Question

Q1))) A student carried out this experiment and found:

mass gelatin capsule = (1.2010x10^-1) grams

mass alloy sample + gelatin capsule = (2.86x10^-1) grams

mass empty beaker= (1.4083x10^2) grams

mass beaker + displaced water= (2.9086x10^2) grams

temperature= 294.15 K

Barometric pressure= 746 mmHg

vapor pressure of water= 18.65 mmHg

How many liters of hydrogen gas were produced? Enter answer in scientific notation with three significant figures.

Q2.))) A student performed this lab, but accidentally used the barometric pressure in his calculations instead of the pressure of dry hydrogen. How would this have changed his calculated moles of hydrogen gas and calculated mass of zinc in the alloy?

1.-- moles of hydrogen would decrease and grams of zinc would increase

2.-- moles of hydrogen would increase and grams of zinc would decrease.

3.-- moles of hydrogen would decrease and grams of zinc would decrease

Answer 1

Part a

Mass of water displaced = (mass beaker + displaced water)-( mass empty beaker)

= 2.9086x10^2 - 1.4083x10^2

= 150.03 g

Volume of water displaced = volume of H2 produced

= mass of water/density of water

= 150.03g/1g/mL

= 150.03 mL

Part b

If barometric pressure is used

Moles of H2

= (746/760)atm x 0.150 L / 0.0821 L-atm/mol-K x 294.15 K

= 0.0061 mol

The balanced reaction

Zn + 2HCl = ZnCl2 + H2

Moles of Zn consumed = 0.0061 mol

Mass of Zn = moles x molecular weight

= 0.0061 mol x 65.38g/mol

= 0.398 g

If dry pressure is used

Moles of H2

= (746-18.65/760)atm x 0.150 L / 0.0821 L-atm/mol-K x 294.15 K

= 0.0059 mol

The balanced reaction

Zn + 2HCl = ZnCl2 + H2

Moles of Zn consumed = 0.0059 mol

Mass of Zn = moles x molecular weight

= 0.0059 mol x 65.38g/mol

= 0.388 g

On using barometric pressure instead of dry pressure

Moles of H2 increases and mass of Zn also increased