A student carefully placed 23.0 g of sodium in a reactor with an excess quantity of chlorine gas. When the reaction is complete, the student obtained 58 grams of salt. How many grams of sodium reacted?

Respuesta :

Answer: 22.8 grams

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} NaCl=\frac{58g}{58.5g/mol}=0.99moles[/tex]

The balanced chemical reaction is :

[tex]2Na+Cl_2\rightarrow 2NaCl[/tex]

According to stoichiometry :

As 2 moles of [tex]NaCl[/tex] is produced by = 2 moles of [tex]Na[/tex]

Thus 0.99 moles of [tex]NaCl[/tex] is produced by =[tex]\frac{2}{2}\times 0.99=0.99moles[/tex]  of [tex]Na[/tex]

Mass of [tex]Na=moles\times {\text {Molar mass}}=0.99moles\times 23g/mol=22.8g[/tex]

Thus 22.8 g of [tex]Na[/tex] is reacted.