Question

For the following metal and non-metal combination reaction you have 3 moles of Na. How many moles of Cl2 would you need to add to have stoichiometrically equivalent amounts of NA and Cl2?

2Na(s) + Cl2(g) --> 2NaCl(s)

show work so i can understand please.

Answer 1

First of all, moles to moles conversions need a balanced chemical equation (future reminder if you haven't already done so). In this case the formula is balanced, so no need to worry about that.

Equation for moles to moles conversion:
# of moles of substance * ratio of substance
_________________ = # of moles of the substance you need
ratio of substance
Here we go,

3 moles of Na * 2Na
---- = 3 moles of Cl
2Cl
Explanation:
From what you gave me, there were 3 moles of Na in that equation you gave me, so I plugged it into the equation for moles to moles conversion. From the balanced chemical equation I got 2 Na. You might be asking, why 2 Na? Simple, there are 2 Na molecules on both sides of the equation; there are also 2 Cl molecules on both sides. Plug that into the equation and you get 2Na/2Cl.

Now, a common mistake when calculating moles to moles is that the 2Na/2Cl (substance to substance ratio) is often switched around. Make sure you put the # of molecules of the substance you want to find at the bottom. In this case you want to find the # moles of Cl, so it goes at the bottom (hence, 2Na/2Cl.

Back to solving the equation. Multiply 3*2. The answer is 6, and 6/ 2= 3. The final answer is 3 moles of Cl.

Answer 2

Actually, you would only need 1.5 moles of Cl2.

As you can see in the stoichiometrically balanced equation given, the between sodium metal (Na) and chlorine gas (Cl2) requires two molar equivilants of the metal for every mole of the gas. Thus, if you need 3 moles of Na, you need 1.5 moles of Cl2.