Answer to Question #181004 in General Chemistry for Sophia Beer

Question #181004

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
2021-04-19T05:11:08-0400

no. of mole of Fe = "\\frac{6.5\u00d710^{23}}{6.022\u00d710^{33}}" = 1.08 mol


for 3 mol of Fe = 2 mol of Au(NO3)3

1.08 mol of Fe = "\\frac{2}{3} \u00d7 1.08"

= 0.72 mol

1 M = (1 mol / 1 L)

1 mol = 1 L

0.72 mol = 0.72 L


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