A sample of sodium reacts completely with 142 g of chlorine, forming 234 g of sodium chloride. What mass of sodium reacted?
I took 142g of Cl / 35.5g/mol = 4 mols of Cl
Then I took 58.44g/mol x 4 mols of Cl = 233.76
It asked for three significant figures only so I put 233 and 234 but i got the worng answer twice.
Somebody plz help!
Write out your balanced equation for the reaction first.
2Na + Cl2 --> 2NaCl
? 142g 234g
Mols of Cl2
142g / 71 g/mole
= 2 mol of Cl2
Because of the balanced equation, one mol of Cl2 requires 2 mols of Na for all of the chlorine to be used up
... 4 mol of Na
4 mol * 23 g/mol
= 92 g of Na
The other way to do this...
234g sodium chloride - 142g of chlorine = 92g of sodium
NaCl = 23+35.5 = 58.5 g/mol
so 234/58.5 = 4 mols of NaCl
that requires
23 * 4 = 93 grams of Na
To determine the mass of sodium that reacted, we need to use stoichiometry, which is the relationship between the moles of reactants and products in a chemical reaction.
First, let's calculate the moles of sodium chloride formed:
Mass of sodium chloride = 234 g
Molar mass of sodium chloride (NaCl) = 58.44 g/mol (which you correctly used)
Moles of sodium chloride = Mass / Molar mass = 234 g / 58.44 g/mol = 4 moles
According to the balanced chemical equation for the reaction between sodium and chlorine, the ratio of moles of sodium to sodium chloride is 2:1:
2 Na + Cl2 -> 2 NaCl
So, if we have 4 moles of sodium chloride formed, we would expect 2 moles of sodium to have reacted.
Therefore, the amount of sodium that reacted can be calculated as follows:
Moles of sodium reacted = 2 moles
Finally, we can use the molar mass of sodium (22.99 g/mol) to convert the moles of sodium into the mass of sodium:
Mass of sodium reacted = Moles of sodium reacted x Molar mass of sodium
Mass of sodium reacted = 2 moles x 22.99 g/mol = 45.98 g
Hence, the mass of sodium that reacted is 45.98 g.