if 14.7g of na reacts with 7.9g og br how many grams of nabr are found
convert to moles
14.7g Na = 0.639 moles Na
7.9g Br = 0.099 moles Br
Since each mole of Na reacts with 1 mole of Br, you should be able to figure out the result. (There will be some Na left over.)
Anna, you can't get lazy in chemistry and not use caps when needed. na is not anything but two letters and I have no idea what og br is. Starting a sentence with a capital letter at least tells us where the sentence starts. A period helps us know where the sentence ends.
This is not a tweet or facebook board and if you want help it behooves you to post the question with as much clarity as possible. Steve probably is smarter than I; he may have figures out that og is a typo and you intended to write of.
To determine the number of grams of NaBr formed when Na reacts with Br, we need to first calculate the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
We can find the limiting reactant by comparing the ratio of the reactants to the stoichiometric coefficients in the balanced equation for the reaction.
The balanced equation for the reaction between Na and Br is:
2Na + Br₂ -> 2NaBr
The molar mass of Na is 22.99 g/mol, and the molar mass of Br₂ is 159.81 g/mol.
To find the number of moles of each reactant, we can use the equation:
moles = mass / molar mass
Moles of Na = 14.7 g / 22.99 g/mol ≈ 0.639 mol
Moles of Br₂ = 7.9 g / 159.81 g/mol ≈ 0.049 mol
Now, let's compare the mole ratios of Na and Br₂ in the balanced equation:
Na:Br₂ = 2:1
The mole ratio tells us that for every two moles of Na, we need one mole of Br₂ to react completely.
Since the mole ratio is 2:1, and we have less moles of Br₂ (0.049 mol) compared to Na (0.639 mol), Br₂ is the limiting reactant.
To calculate the amount of NaBr formed, we need to find the number of moles of NaBr using the mole ratio from the balanced equation:
moles of NaBr = moles of Br₂ × (2 moles of NaBr / 1 mole of Br₂)
= 0.049 mol × (2 moles of NaBr / 1 mole of Br₂)
= 0.098 mol
Finally, we can find the mass of NaBr formed by multiplying the number of moles of NaBr by its molar mass:
mass of NaBr = moles of NaBr × molar mass of NaBr
= 0.098 mol × (22.99 g/mol + 79.9 g/mol)
≈ 10.6 g
Therefore, approximately 10.6 grams of NaBr are formed.