Answer to Question #180684 in General Chemistry for Enci Peng

Question #180684

Throwing some scrap iron in a gold nitrate solution causes the gold metal to precipitate. How many liters of 1 M gold nitrate solution would react with 6.5x1023 particles of iron metal?

3 Fe(s) + 2 Au(NO3)3(aq) --> 3 Fe(NO3)2(aq) + 2 Au(s)



1
Expert's answer
2021-04-14T06:10:27-0400

From the equation

2 moles of Au(NO3)3(aq) = 3 moles of Fe

= 6.5 ×1023/6.022 × 1023 = 1.0794 moles

= 1.0794 × 2/3 = 0.7196

Molar Mass of gold nitrate = 382.98

= 0.7196 × 382.98

= 275.59g

= 0.2756 L



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