a 3.00mole sample of fluorine reacts with sodium bromide to produce how many grams of sodium fluoride?
Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine how many grams of sodium fluoride will be produced when a 3.00 mole sample of fluorine reacts with sodium bromide, we need to use the mole ratio from the balanced chemical equation.
The balanced chemical equation for the reaction between fluorine (F2) and sodium bromide (NaBr) is:
2 F2 + 2 NaBr -> 2 NaF + Br2
According to this equation, 2 moles of fluorine (F2) react with 2 moles of sodium bromide (NaBr) to produce 2 moles of sodium fluoride (NaF).
Therefore, the mole ratio of F2 to NaF is 2:2, which simplifies to 1:1. This means that for every 1 mole of F2 reacting, 1 mole of NaF is produced.
Now, we can use the given information that we have a 3.00 mole sample of F2 to find the number of moles of NaF produced. Since the mole ratio is 1:1, we know that 1 mole of NaF will be produced for every 1 mole of F2.
Therefore, the number of moles of NaF produced is also 3.00 moles.
To find the grams of NaF produced, we need to know the molar mass of NaF. The molar mass of sodium (Na) is approximately 22.99 g/mol, and the molar mass of fluorine (F) is approximately 18.99 g/mol. Adding these values together, we get the molar mass of NaF:
Molar mass of NaF = 22.99 g/mol + 18.99 g/mol = 41.98 g/mol
Now, we can convert the moles of NaF to grams using the molar mass:
Number of grams of NaF = Number of moles of NaF × Molar mass of NaF
Number of grams of NaF = 3.00 moles × 41.98 g/mol
Calculating this, we get:
Number of grams of NaF = 125.94 g
Therefore, when a 3.00 mole sample of fluorine reacts with sodium bromide, it will produce 125.94 grams of sodium fluoride.