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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)

Sagot :

Azkaaidn

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 [tex]\frac{2}{3}[/tex] = 1.2343... x [tex]\frac{2}{3}[/tex] = 0.82287...

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

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