For the chemical reaction na 2 co 3 + ca ( no 3 ) 2 ⟶ caco 3 + 2 nano 3 how many moles of calcium carbonate will be produced from 91.9 g of sodium carbonate? moles of calcium carbonate:

Na2CO3 mol = mass/rfm
= 91.9 / ( 23+23+12+16+16+16)
= 91.9/106
=919/1060 or roughly 0.867

Mole ratio of sodium carbonate to calcium carbonate is 1:1

Therefore calcium carbonate mol = mass/rfm

919/1060 = mass / (40+12+16+16+16)
919/1060 = mass / 100
Mass = 919/1060 x 100
= 4595/53 or 86.7g to 3 significant figures

The answer is b on plato.