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

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} Na=(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 =
`(2)/(2)* 2.56=2.56moles` of NaCl

Thus 2.56 moles of NaCl are produced.