Answer
There is a total of $15.9$ g in 0.150 mole of $Na_2CO_3$
Work Step by Step
1. Determine the molar mass of this compound ($Na_2CO_3$), and setup the conversion factors:
Molar mass :
$Na: 22.99g * 2= 45.98g $
$C: 12.01g $
$O: 16.00g * 3= 48.00g $
45.98g + 12.01g + 48.00g = 105.99g
$ \frac{1 \space mole \space (Na_2CO_3)}{ 105.99 \space g \space (Na_2CO_3)}$ and $ \frac{ 105.99 \space g \space (Na_2CO_3)}{1 \space mole \space (Na_2CO_3)}$
2. Calculate the mass:
$ 0.150 \space mole \times \frac{ 105.99 \space g}{1 \space mole} = 15.9 \space g$
