Answer:
[tex]m_{NaHCO_3}=16.8gNaHCO_3[/tex]
Explanation:
Hello!
In this case, since the molecular formula of sodium bicarbonate is:
[tex]NaHCO_3[/tex]
We realize it has a molar mass of 84.007 g/mol; therefore, as we need the mass in 0.200 moles of sodium bicarbonate, we develop the following setup:
[tex]m_{NaHCO_3}=0.200NaHCO_3*\frac{84.007gNaHCO_3}{1molNaHCO_3}[/tex]
Thus, we obtain:
[tex]m_{NaHCO_3}=16.8gNaHCO_3[/tex]
Best regards!
