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GrogVix [38]
2 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]2 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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Explanation:

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3 years ago
How to do q solution, qrxn, moles of Mg , and delta Hrxn?
Helga [31]

Answer:

<em> 14, 508J/K</em>

ΔHrxn =q/n

where q = heat absorbed and n = moles

Explanation:

<em>m = mass of substance (g) = 0.1184g</em>

1 mole of Mg - 24g

<em>n</em> moles - 0.1184g

<em>n = 0.0049 moles.</em>

Also, q = m × c × ΔT

<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>

<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>

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ΔT=  (final - initial) temp = 38.3 - 27.2

= 11.1 °C.

mass of MgCl2 = 95.211 × 0.1184 = 11.27

⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>

<em>= 1,7117.7472 J °C-1 g-1</em>

<em />

<em>∴ ΔHrxn = q/n</em>

<em>=1,7117.7472  ÷ 0.1184 </em>

<em>= 14, 508J/K</em>

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