Question

given the equation 2Na+Cl2 ​ →2NaCl How many moles of NaCl will be produced from 59.0 grams of Na, assuming Cl is available in excess?

Answer 1

I would use dimensional analysis for this problem. You would start with the given amount of Na which is 59.0g. You are trying to find the moles of NaCl.

(59.0g Na)*(1 mol Na)/(22.99 g Na)*(2 mol NaCl)/(2 mol Na)=

2.57 mol NaCl

The units will cancel out until you are left with moles of NaCl. The answer has the correct number of significant figures (3 sig figs).

Answer 2

Answer: 2.56 moles

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} Na=\frac{59.0g}{23g/mol}=2.56moles$

$2Na(s)+Cl(g)\rightarrow 2NaCl(s)$

Na is the limiting reagent as it limits the formation of product and chlorine is the excess reagent.

According to stoichiometry :

2 moles of Na produce = 2 moles of NaCl

Thus 2.56 moles of Na produce =$\frac{2}{2}\times 2.56=2.56moles$  of NaCl

Thus 2.56 moles of NaCl are produced.