Question

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`?

Answer 1

`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