# Question #c7ba8

Feb 29, 2016

$\approx 38.6 g m$ rounded to first place of decimal.

#### Explanation:

Given reaction is
$2 \text{Na" + "Cl"_2 -> 2"NaCl}$

We need to consider reactants as weight of sodium is given and we are required to ascertain the weight of chlorine needed.
Now average of Atomic mass of Chlorine $= 35.4527$ amu and of Sodium$= 22.98977$ amu

From the balanced equation we see that 2 atoms of sodium are required for one molecule of chlorine.
Or one atom sodium combine with one atom of chlorine.

$\therefore$ weight of chlorine needed for $22.98977 g m$ Sodium$= 35.4527 g m$
weight of chlorine needed for $25 g m$ Sodium $= \frac{35.4527}{22.98977} \times 25$
$\approx 38.6 g m$ rounded to first place of decimal.

Assume that:
n - number of moles
m - mass of substance
M - molar mass (or atomic weight on periodic table)
$n = m \div M$

Since 25 grams (m) of Na has been given to you, you need to find the molar mass (M) of Na.

If you look into your periodic table, the molar mass (M) of Na is 23.0 g/mol.

Now you have to find the number of moles (n) in Na.
The number of moles in sodium is:
[n = 25 grams $\div$ 23.0 g/mol] = 1.086956522 moles.
Note: At this stage, don't round off because this is not final answer, otherwise you would get a slightly different answer.

The next step is to find the number of moles (n) for Cl. First, you should know the mole ratio between Na and Cl. The mole ratio of Na:Cl is 2:1.
If 2 moles of Na gives you 1 mole of Cl, then 1.086956522 moles of Na should give you: [1.086956522 $\div$ 2 $\times$ 1 ]
= 0.5434782609 moles of Cl. Now you know how many moles (n) in Cl.

Now find the molar mass (M) of Cl. If you refer to your periodic table again, the molar mass of Cl is 35.5 g/mol.

The final step is to find the mass (m) of Cl.
$m = M \times n$.
The mass of Cl is: [m = 35.5 $\times$0.5434782609] = 19.3 grams.
Note: The final answer is rounded to 3 significant figures.

Therefore, 19.3 grams of Cl would be needed if you were given 25 grams of Na.