Chemistry Stoichiometry

Question

If in the double replacement reaction of excess iron (II) phosphate with 23.79 kilograms of aluminum nitrate the products are iron (II) nitrite and aluminum phosphate, how many particles of iron (II) nitrate are produced? Must be in scientific notation.

Answer 1

First we should write out the entire unbalanced equation:

__Fe₃(PO₄)₂ + __Al(NO₃)₃ --> __Fe(NO₃)₂ + __Al(PO₄)

The first step would be to balance this equation. We can see that both iron only appears once on either side of the equation so we can say the 3 iron(ii) ions get carried over to iron(ii) nitrate so we can write 1 in front of Fe₃(PO₄)₂ and 3 in front of Fe(NO₃)₂. Since we have 2 phosphates on the reactants side, we can then write a 2 in front of Al(PO₄) on the products side. Finally, we can see that 6 nitrates are generated in the products so we need 6 nitrates on the reactants side. We can add a 2 in front of Al(NO₃)₃.

Fe₃(PO₄)₂ + 2 Al(NO₃)₃ --> 3 Fe(NO₃)₂ + 2 Al(PO₄)

Since we know aluminum nitrate is limiting, we can calculate the number of moles that will be transformed into products.

23.79 (kg) / 213 (g/mol) *1000 (g/kg) = 111.7 moles

Since 2 moles of aluminum nitrate have to react to produce 3 moles of iron(ii) nitrate we can then expect 167.5 moles of iron(ii) nitrate to be produced. Now it's unclear what you mean by particles, but to find the number of molecules of iron(ii) nitrate, simply multiply this by Avogadro's number to obtain the answer.

167.5 * 6.23x10²³ = 1.04x10²⁶ molecules iron(ii) nitrate