What are the grams of sodium chloride produced when 10.0 g of sodium react with 10.0 g of chlorine gas in the equation #2Na+Cl_2 -> 2NaCl#?

1 Answer
Aug 16, 2017

Answer:

#m (NaCl)=16.479#g

Explanation:

1) We need to know which of reagents is excess,
so we calculate it by reagents' mass.
#M (Na)=23# #frac\{g}{mol}#
#m (Na)=n×M=2×23=46# g

#M (Cl_2)=35.5×2=71# #frac\{g}{mol}#
#m (Cl_2)=n×M=1×71=71# g

#\frac{10}{46}:\frac{x}{71}#

#x=\frac {10×71}{46}=15.43#

This means the sodium is excess.
So, we're continuing solving the problem by chlorine.

2) #M (NaCl)=23+35.5=58.5# #frac\{g}{mol}#
#m (NaCl)=n×M=2×58.5=117#g

3) #10g (Cl_2) - xg (NaCl)#
#71g (Cl_2) - 117g (NaCl)#

#x=\frac {10×117}{71}=16.479# g