Question

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 bromine combined with unlimited sodium. using correct significant figures.

Answer 1

I came up with 5.72 g Br2.

Answer 2

It's asking for amount of salt (NaBr), not \(Br_2\), always start by writing a chemical equation for the process.

Answer 3

Here's a tip that applies to just about every field: when you think you have an answer to a question, reread the question and ask yourself if your proposed answer answers the question asked.

Here, the question asks "how many grams of salt can be made?" and your answer was 5.72 g Br2. Is that answering the question? No.

Even when your answer does answer the right question, you should also stop and assess the reasonability of your answer.

Question: how many grams of salt can be made from 5.00 grams of bromine

Your answer: 5.72 g Br2

How did you transform 5.00 grams of Br into 5.72 grams of Br in a chemical reaction? If this is for real, we can make a lot of money :-)

Answer 4

I actually meant 5.72 g NaBr, but even that is wrong.

Here's my work:

2Na + Br2 = 2NaBr
5 g Br2 (1 mol Br2/159.81 g Br2) (1 mol Br2/2 mol NaBr) (182.8 g NaBr/1 mol Br2)

That's 2.86 g NaBr

Answer 5

The mole ratio is incorrect, the moles of Br2 should be multiplied by 2, not divided.

Answer 6

How should it be written?

Answer 7

write the reciprocal of what you wrote

Answer 8

5 g Br2 (1 mol Br2/159.81 g Br2) (2 mol NaBr/1 mol Br2) (182.8 g NaBr/1 mol Br2)

= 11.44 g NaBr

Answer 9

This is an error easily prevented by actually USING the units you wrote.

\[5 \text{ g Br}_2* (\frac{1 \text{ mol Br}_2}{159.81\text{ g Br}_2}) (\frac{1 \text{ mol Br}_2}{2\text{ mol NaBr}}) (\frac{182.8\text{ g NaBr}}{1\text{ mol Br}_2})\]
now carefully cancel units:
\[5 \cancel{\text{ g Br}}_2* (\frac{1 \text{ mol Br}_2}{159.81\cancel{\text{ g Br}_2}}) (\frac{1 \text{ mol Br}_2}{2\text{ mol NaBr}}) (\frac{182.8\text{ g NaBr}}{1\text{ mol Br}_2})\]\]\[5 \cancel{\text{ g Br}}_2* (\frac{1 \text{ mol Br}_2}{159.81\cancel{\text{ g Br}_2}}) (\frac{1 \cancel{\text{ mol Br}_2}}{2\text{ mol NaBr}}) (\frac{182.8\text{ g NaBr}}{1 \cancel{\text{ mol Br}_2}})\]\]

Uh, no way that you end up with just \(\text{g NaBr}\) here, right? That means you have an incorrect equation.

With the correct equation:
\[5 \text{ g Br}_2* (\frac{1 \text{ mol Br}_2}{159.81\text{ g Br}_2}) (\frac{2\text{ mol NaBr}}{1 \text{ mol Br}_2}) (\frac{182.8\text{ g NaBr}}{1\text{ mol NaBr}})\]\[5 \cancel{\text{ g Br}_2}* (\frac{1 \text{ mol Br}_2}{159.81\cancel{\text{ g Br}_2}}) (\frac{2\text{ mol NaBr}}{1 \text{ mol Br}_2}) (\frac{182.8\text{ g NaBr}}{1\text{ mol NaBr}})\]\[5 \cancel{\text{ g Br}_2}* (\frac{1 \cancel{\text{ mol Br}_2}}{159.81\cancel{\text{ g Br}_2}}) (\frac{2\text{ mol NaBr}}{1 \cancel{\text{ mol Br}_2}}) (\frac{182.8\text{ g NaBr}}{1\text{ mol NaBr}})\]\[5 \cancel{\text{ g Br}_2}* (\frac{1 \cancel{\text{ mol Br}_2}}{159.81\cancel{\text{ g Br}_2}}) (\frac{2\cancel{\text{ mol NaBr}}}{1 \cancel{\text{ mol Br}_2}}) (\frac{182.8\text{ g NaBr}}{1\cancel{\text{ mol NaBr}}})\]\[5 \cancel{\text{ g Br}_2}* (\frac{1 \cancel{\text{ mol Br}_2}}{159.81\cancel{\text{ g Br}_2}}) (\frac{2\cancel{\text{ mol NaBr}}}{1 \cancel{\text{ mol Br}_2}}) (\frac{182.8\text{ g NaBr}}{1\cancel{\text{ mol NaBr}}})\]\[= \frac{5*2*182.8}{159.81*1*1}\text{ g NaBr} =11.44 \text{ g NaBr} \]