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 $2Na+Cl_2 -> 2NaCl$ ?

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Answer 1 · 2017-08-16T08:29:51

$m (NaCl)=16.479$ g

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

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 -> 2NaCl2Na+Cl_2 -> 2NaCl?