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3 years ago

If 4.20 mol Al was mixed with 1.75 mol Fe2 O3 which reactant is the limiting reactant?

1 answer:

7 0

4.20 mol Al would react completely with 4.20 x (1/2) = 2.10 mol Fe2O3, but there is not that much Fe2O3 present, so Fe2O3 is the limiting reactant. (1.75 mol Fe2O3) x (2/1) x ( 55.8452 g Fe/mol) = 195 g Fe 3 MgO + 2 H3PO4 → Mg3(PO4)2 + 3 H2O (15.0 g MgO) / (40.3045 g MgO/mol) = 0.37217 mol MgO (18.5 g H3PO4) / (97.9953 g H3PO4/mol) = 0.18878 mol H3PO4 0.18878 mol H3PO4 would react completely with 0.18878 x (3/2) = 0.28317 mole of MgO, but there is more MgO present than that, so MgO is in excess and H3PO4 is the limiting reactant. Now we must consider why the problem tells us "17.6g of Mg3(PO4)2 is obtained". The first possibility is that it's just there for the sake of confusion -- in which case ignore it and proceed this way: ((0.37217 mol MgO initially) - (0.28317 mole MgO reacted)) x (40.3045 g MgO/mol) = 3.59 g MgO left over However, if the amount of magnesium phosphate obtained is given because the reaction was stopped before it was complete, the amount obtained governs the amount reacted and the amount left over, so proceed this way: (17.6g Mg3(PO4)2) / (262.8581 g Mg3(PO4)2/mol) x (3/1) = 0.20087 mol MgO reacted ((0.37217 mol MgO initially) - (0.20087 mole MgO reacted)) x (40.3045 g MgO/mol) = 6.90 MgO left over

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

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3 years ago

Why do different materials have similar properties

Deffense [45]

Answer:

The difference in the number of protons and neutrons in atoms account for many of the different properties of elements.

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2 years ago

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How many moles are in 54.0 grams of Silver?

Lelechka [254]

54g ag *(108mol ag/1 g ag) =5832mol ag

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3 years ago

Find the no of moles of ca in 8g of ca

Tom [10]

Answer:

Thus,there are 0.2 mole in 8 g of calcium.

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2 years ago

what term describes when a solvent is holding as much solute in solution as it is able for a given temperature

Serggg [28]

The term that describes when a solvent is holding as much solute in solution as it is able for a given temperature is called supersaturated solution. A little disturbance to this solution some of the solute will crystallize or precipitate out since this solution is not stable.

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