Question

A reaction vessel contains 50.0 grams of sodium metal and chlorine gas. How many grams of chlorine gas are needed to produce 75.0 grams of sodium chloride? If this amount of chlorine gas is added to the vessel, how much sodium metal would remain after the reaction goes to completion?

Answer 1

Answer:

- $m_{Cl_2}=45.5gCl_2$

- $m_{Na}^{unreacted}=20.5g$

Explanation:

Hello,

In this case, we consider the undergoing chemical reaction as:

$2Na+Cl_2\rightarrow 2NaCl$

Thus, for 75.0 grams of sodium chloride, the following grams of chlorine are required (consider their 2:1 molar ratio):

$m_{Cl_2}=75.0gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{1molCl_2}{2molNaCl}*\frac{70.9gCl_2}{1molCl_2} \\ \\m_{Cl_2}=45.5gCl_2$

Then, for the unreacted grams of sodium, we first compute the actually reacted grams by considering the 2:2 molar ratio between sodium chloride and sodium:

$m_{Na}=75.0gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNa}{2molNaCl}*\frac{23.0gNa}{1molNa} \\ \\m_{Na}=29.5gNa$

Finally, we subtract:

$m_{Na}^{unreacted}=50.0g-29.5g\\\\m_{Na}^{unreacted}=20.5g$

Regards.