Create a balanced equation for the reaction between sodium and bromine. Find how many grams of salt can be made from 5.00 grams of sodium and unlimited bromine.
Is it about 39.8 g NaBr2?
To create a balanced equation for the reaction between sodium and bromine, we first need to know the chemical formulas of the reactants and the product. Sodium is represented by the symbol Na, and bromine is represented by the symbol Br2 (since bromine naturally exists as a diatomic molecule).
The balanced equation for the reaction between sodium and bromine is:
2Na + Br2 -> 2NaBr
Now let's calculate how many grams of salt (NaBr) can be made from 5.00 grams of sodium and unlimited bromine.
To do this, we need to convert the given mass of sodium to moles, then use the stoichiometry of the balanced equation to determine the moles of NaBr formed, and finally convert that back to grams.
The molar mass of sodium (Na) is approximately 23 g/mol.
First, let's convert the grams of sodium (Na) to moles:
5.00 g Na * (1 mol Na / 23.0 g Na) = 0.217 mol Na
Since the balanced equation tells us that 2 moles of Na produce 2 moles of NaBr, we can see that the moles of NaBr formed will be the same as the moles of Na used in the reaction.
0.217 mol Na * (2 mol NaBr / 2 mol Na) = 0.217 mol NaBr
Finally, let's convert the moles of NaBr to grams using its molar mass. The molar mass of NaBr is approximately 102.9 g/mol.
0.217 mol NaBr * (102.9 g NaBr / 1 mol NaBr) = 22.3 g NaBr
Therefore, approximately 22.3 grams of NaBr can be made from 5.00 grams of sodium and unlimited bromine, not 39.8 grams as you suggested.