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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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Answer:

The endpoint volume is 50.52 ±  0.14 mL

Explanation:

In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:

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