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9->moles Data Table - Mg with HCI nga2 + H₂ Ptoral - Plzo =PHz Trial 1 Trial 2 Mass of magnesium Atmospheric pressure 0319 76
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Answer #1

Pressue and Temperature are same for both traails

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Mg+2HCl\rightarrow H_{2}(g)+MgCl_{2}

1 mole of Mg liberates 1 mole of Hydrogen gas.

24 g of Mg liberates 2 g of hydrogen gas in mass terms.

0.031 g of Mg liberates how much hydrogen gas= ?

Mass of hydrogen released in Trail 1 = ( 0.031 g * 2 g ) / 24 g

= 0.00258 g

Moles of Hydrogen gas = 0.00258 g/ 2 g/mol

= 0.00129 mol of Hydrogen

value of Gas constant, R is derived by Ideal gas equation -----

pV = nRT

R = pV/nT

T = 293.25 K ( 20.1 ^oC +273.15)

pressure of dry hydrogen gas , P = 767.4 mm- 17.5 mm

= 749.9 mm

=( 749.9/760 ) atm

= 0.987atm

volume of gas = 30.70 ml

= 0.0307 L

------------------------------------------------------

R = pV/nT

= 0.987 atm * 0.0307 L ) / 0.00129 mol * 293.25 K

= 0.080 Lit.atm/K.mol --------------------------------------------------------------------trail 1

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1 mole of Mg liberates 1 mole of Hydrogen gas.

24 g of Mg liberates 2 g of hydrogen gas in mass terms.

0.042 g of Mg liberates how much hydrogen gas= ?

Mass of hydrogen released in Trail 2 = ( 0.042 g * 2 g ) / 24 g

= 0.0035 g

Moles of Hydrogen gas = 0.0035 g/ 2 g/mol

= 0.00175 mol of Hydrogen

value of Gas constant, R is derived by Ideal gas equation -----

pV = nRT

R = pV/nT

T = 293.25 K ( 20.1 ^oC +273.15)

pressure of dry hydrogen gas , P = 767.4 mm- 17.5 mm

= 749.9 mm

=( 749.9/760 ) atm

= 0.987atm

volume of gas = 43.05 ml ml

= 0.04305 L

------------------------------------------------------

R = pV/nT

= 0.987 atm * 0.04305 L ) / 0.00175 mol * 293.25 K

= 0.0828 Lit.atm/K.mol -------------------------------------------------------------------------trail 2

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