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Data Sheet Unknown Number Sample 2 Sample 1 Volume of Ha gas (ml.) 22.65 22.12 Temperature of water (C) 20 8 0 20.8 C Height
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Answer #1

sample 1

vapor pressure of water at the given temperature = 18.7 mmHg

vapor pressure of dry hydrogen = total pressure - water vapor pressure.

= 747.776 mmHg - 18.7 mmHg

= 729.976 mmHg

temperature = 20.8 + 273 k

= 293.8K

R = 0.0821lit.atm/k.mol

volume of gas = 22.65 ml

= 0.02265 lit

number of moles of hydrogen = pV/RT

= 729.976 mmHg x 0.02265 lit / 0.0821lit.atm/k.mol x 293.8K

= 0.6846 moles

mass of magnesium ribbon

1 equivalent of magnesium metal = 1 equivalent of hydrogen gas

so numbe rof moles of Mg = moles of hydrogen

= 0.6846 moles

mass of Mg ribbon = 0.6846 moles x 24 g/mol

= 16.43 g

sample2 ----- same Temperature

vapor pressure of water at the given temperature = 18.7 mmHg

vapor pressure of dry hydrogen = total pressure - water vapor pressure.

= 747.776 mmHg - 18.7 mmHg

= 729.976 mmHg

temperature = 20.8 + 273 k

= 293.8K

R = 0.0821lit.atm/k.mol

volume of gas = 22.12 ml

= 0.02212 lit

number of moles of hydrogen = pV/RT

= 729.976 mmHg x 0.02212 lit / 0.0821lit.atm/k.mol x 293.8K

= 0.6694 moles of hydrogen gas = moles of Mg

mass of Mg ribbon = moles of Mg ribbon metal x 24 g/mol

= 0.6694 x 24 g/mol

= 16.0656 g

average mass of Mg ribbon = (16.43 g + 16.0656 g )/2

= 16.2478 g

  

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