Question
4 What mass of sodium fluoride is formed when 2.30 g of sodium reacts with 2.85 g of fluorine?
Answer
4.4
(394 Votes)
Stanley
Master · Tutor for 5 years
Answer
Step 1:The molar mass of sodium = 22.99 g/mol. Hence, the moles of sodium in 2.30 g = 2.30 /? 22.99 = 0.1 mol. Gemini Electron also suggests that the molar mass of fluorine is = 38 g /mol. Hence, the mol of fluorine in 2.85 g is 2.85/38 = around 0.075 mol.Step 2:The reaction is 2Na + F2 → 2NaF So, 2 moles of sodium would react to give 2 moles of sodium fluoride. But, given that sodium is present in excess, every .075 mols of fluorine would react with .075 densities of CH4, giving .075 moles of NaF Step 3: =Formatting .075 moles of NaF, as stated in step 2, must also be converted back to mass to reach the final answer. Hence, using the molar mass of NaF ( 41.99): .075 mol of NaF would be = max I = quantity("41.99 g NaF /mol")shoma I= *.075 mols = autour ( 3.15 g of NaF).As an educated guess, the format of sodium fluoride produced is roughly 7 g."
Explanation
This is a stoichiometry problem involving chemical reactions, and could be tackled leveraging the law of conservation of mass and the molecular weights to find the answer. For sodium (Na) it's atomic mass is around 22.99 g/mol and for fluorine (F2) it has a molecular mass of round 38 g/mol. Combining these two in a reaction would yield Sodium Fluoride (NaF) which has a molecular mass of around 41.99 g/mol.Given that the reaction is 2Na + F2 -> 2NaF; it's clear that for each 1 Mol of Sodium we would get 1 Mol of Sodium Fluoride. But, the mass of Sodium is given not in moles, but in grams. Hence, first we would convert the mass given in grams to moils to get our answer.Finally, the masses given in the question would not ideally exhaust each other and hence we must determine the limiting reagent to correctly determine which mass of sodium flouride is yielded.