Answer:0.119g
Explanation:equation of rxn is
2Al+6HCl=2AlCl3+3H2
From ideal gas eqn
PV=nRT
n=PV/RT
P here is the partial pressure of H2 from the qtn.According to Dalton law of partial pressure, PT=PH2+PH20
PT=0.89atm given
PH20=25.2mmhg given=25.2/760atm,=0.033atm
PH2=PT-PH20
PH2=0.89-0.033=0.857atm
T=26+273=299K
R=0.082atmdm^-3mol^-1K^-1
V=190.6ml=190.6cm3=190.6/1000=0.1906dm3
n=PV/RT
n=0.857*0.1906/0.082*299
=0.00667moles of H2.
From the eqn of reaction,
2moles of Al reacts to gv 3moles of H2
xmoles of Al will give 0.00667moles of H2
xmoles=0.00667*2/3 (cross multiplying)=0.00444moles of Al
From the relationship, n=mass/MW
mass=MW*n
MW of Al=27g/mol
mass=0.0044moles*27g/moles
mass=0.119grams of Al.
The mass of aluminum that has reacted is 0.119 g. The mass of the reactant can be calculated by finding its moles.
the mass of the reactant can be calculated by finding its moles in the reaction and putting the value in the mole formula.
The given reaction is:
[tex]\bold {2Al+6HCl\rightarrow 2AlCl_3+3H_2}[/tex]
First, calculate the moles of the Hydrogen from the ideal gas equation,
[tex]n=\dfrac {0.857\times 0.1906}{0.082\times 299}\\\\ n = \rm 0.00667 \ moles \ of \ H2.[/tex]
The molar ratio between Al and [tex]\bold {H_2 }[/tex] is 3:2.
Thus, moles of Aluminium is:
[tex]\text{Moles of Al} = 0.00667\times \dfrac 23 \\\\\text{Moles of Al} = 0.00444 \rm \ moles[/tex]
Thus the mass of Aluminium,
[tex]m ={\rm 0.0044 \ moles\times 27 \ g/moles}\\\\m = 0.119\rm \ g[/tex]
Therefore, the mass of aluminum that has reacted is 0.119 g.
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