How many grams of lithium carbonate are produced from 0.51 moles of lithium phosphate and excess iron(II) carbonate? 3 FeCO3 + 2 Li3PO4 --> 3 Li2CO3 + Fe3(PO4)2

Question

Don't know how to do this.

Answer 1

mol to mol with coefficients

0.51 mol Li3PO4 (3 mol Li2CO3 / 2 mol Li3PO4)=0.765 mol Li2CO3

mol to g with molar mass. Note that coefficients don't affect molar mass.

0.765 mol Li2CO3 (73.888 g Li2CO3 / 1 mol Li2CO3)=56.524=57 g Li2CO3

Answer 2

Check out a similarly solved problem here.

https://youtu.be/45mKzpJ5iIg