6) Mixing 613.7 mL of 0.816 M Fe(NO), with excess (NH4),CO, solution, how many grams of...

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Answer 1

Answer #1

2 Fe(NO3)3 + 3 (NH4)2CO3 = Fe2(CO3)3(s) + 6 NH4NO3

Moles of Fe(NO3)3 = molarity * volume ( in L )

= 0.816 * ( 613.7*10^-3)

= 0.501

Moles of Fe2(CO3)3 = 1/2 * moles of Fe(NO3)3

= 1/2 * 0.501

= 0.2505

Mass of Fe2(CO3)3 = moles * molar mass

= 0.2505 * 292

= 73.146 grams

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Answer 2

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