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GrogVix [38]
3 years ago
11

How many moles of aluminum (Al) produced in the reaction if 30.00 grams of magnesium (Mg) was consumed. (Atomic mass for Mg =24.

305 g/mol)
2AlPO4 + 3Mg rightwards arrow 2Al + Mg3(PO4)2

0.62 moles of (Al)
0.82 moles of (Al)
0.42 moles of (Al)
Chemistry
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

0.82 moles of (Al)

Explanation:

moles = mass ÷ RFM

therefore moles of Mg in equation:

moles of Mg = 30g ÷ 24.305 = 1.2343...

since the ratio of moles from the Mg to Al is 3:2

moles of Al = moles of Mg x \frac{2}{3} = 1.2343... x \frac{2}{3} = 0.82287...

which is rounded down at 2 d.p to 0.82 moles

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A container of N2O3(g) has a pressure of 0.265 atm. When the absolute temperature of the N2O3(g) is tripled, the gas completely
Rzqust [24]

Answer:

1.59 atm

Explanation:

The reaction is:

N_{2}O_{3}(g) - - -> NO_{2}(g)+NO(g)

The dalton's law tell us that the total pressure of a mixture of gases is the sum of the partial pressure of every gas.

So after the reaction the total pressure is:

P_{total}=P_{NO_{2}}+P_{NO}

we don't include N_{2}O_{3} because it decomposed completely.

Assuming  ideal gases

PV=nRT

P= pressure, V= volume of the container, n= mol of gas, R=constant of gases and T=temperature.

so moles of N_{2}O_{3} is:

n_{N_{2}O_{3}}=\frac{P_{1}V}{RT_{1}}

from the  reaction stoichiometry (1:1) we have that after the reaction the number of moles of each product is the same number of moles of N_{2}O_{3}.

n_{NO_{2}}=\frac{P_{1}V}{RT_{1}}

n_{NO}=\frac{P_{1}V}{RT_{1}}

The partial pressure of each gas is:

P_{NO_{2}}=\frac{n_{NO_{2}*R*T_{2}}}{V}

P_{NO}=\frac{n_{NO}*R*T_{2}}{V}

so total pressure is:

P_{total}=(n_{NO_{2}}+n_{NO})*\frac{R*T_{2}}{V}

replacing the moles we get:

P_{total}=(2*\frac{P_{1}V}{RT_{1}})*\frac{R*T_{2}}{V}

We know that T2=3*T1

replacing this value in the equation we get:

P_{total}=2*\frac{P_{1}V}{RT_{1}}*\frac{R*3T_{1}}{V}

P_{total}=6*P_{1} = 6*0.265 atm = 1.59 atm

5 0
3 years ago
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