L35

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Jasmine Lam 1I
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Joined: Fri Sep 25, 2015 3:00 am

L35

Postby Jasmine Lam 1I » Fri Oct 02, 2015 10:22 am

Sodium bromide, NaBr, which is used to produce AgBr for use in photographic film, can itself be prepared as follows.
FeBr+Br2 ---> FeBr2
FeBr2+Br2---> Fe3Br8
Fe3Br8+Na2CO3-->NaBr+CO2+FE3O4
What mass of iron, in kg, is needed to produce 2.50t of NaBr?

Helen T 2G
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

Re: L35

Postby Helen T 2G » Fri Oct 02, 2015 2:43 pm

First, you must convert the 2.50 t of NaBr to grams using the conversion factor 1.0 x 10^6 grams = 1 ton. Then you find the molar mass of NaBr (102.89g) and divide the grams of NaBr by NaBr's molar mass. Next, find the mole ratio between Fe and NaBr which is 3 mol Fe: 8 mol NaBr (don't forget to make sure all the equations are balanced before proceeding with the problem). Lastly, multiply by the molar mass of Fe (55.84g) which will result in 5.09 x 10^5g of Fe which is equivalent to 509 kg of Fe.

Hadji Yono-Cruz 2L
Posts: 64
Joined: Fri Sep 28, 2018 12:26 am

Re: L35

Postby Hadji Yono-Cruz 2L » Thu Oct 04, 2018 12:18 pm

Helen T 2G wrote:First, you must convert the 2.50 t of NaBr to grams using the conversion factor 1.0 x 10^6 grams = 1 ton. Then you find the molar mass of NaBr (102.89g) and divide the grams of NaBr by NaBr's molar mass. Next, find the mole ratio between Fe and NaBr which is 3 mol Fe: 8 mol NaBr (don't forget to make sure all the equations are balanced before proceeding with the problem). Lastly, multiply by the molar mass of Fe (55.84g) which will result in 5.09 x 10^5g of Fe which is equivalent to 509 kg of Fe.


Where did you find the conversion factor to be 1.0 x 10^6 grams = 1 ton? When I converted 1 ton to grams on Google it says it is 907185 grams.

DavidEcheverri3J
Posts: 36
Joined: Fri Sep 28, 2018 12:26 am

Re: L35

Postby DavidEcheverri3J » Mon Oct 08, 2018 3:04 pm

904928548 wrote:
Helen T 2G wrote:First, you must convert the 2.50 t of NaBr to grams using the conversion factor 1.0 x 10^6 grams = 1 ton. Then you find the molar mass of NaBr (102.89g) and divide the grams of NaBr by NaBr's molar mass. Next, find the mole ratio between Fe and NaBr which is 3 mol Fe: 8 mol NaBr (don't forget to make sure all the equations are balanced before proceeding with the problem). Lastly, multiply by the molar mass of Fe (55.84g) which will result in 5.09 x 10^5g of Fe which is equivalent to 509 kg of Fe.


Where did you find the conversion factor to be 1.0 x 10^6 grams = 1 ton? When I converted 1 ton to grams on Google it says it is 907185 grams.


The problem is referring to Metric Tons, not US Tons, since US Tons (the one you Googled) is Imperial System, while Metric Tons are Metric System, which is the system commonly used throughout chemistry. Hope this helped!

lwon Dis2I
Posts: 116
Joined: Wed Sep 30, 2020 9:37 pm

Re: L35

Postby lwon Dis2I » Thu Oct 22, 2020 9:51 pm

Why does this problem give us three chemical reaction equations and not just one?

Zoe Kaiser
Posts: 86
Joined: Fri Sep 29, 2023 11:47 am

Re: L35

Postby Zoe Kaiser » Tue Oct 24, 2023 10:13 pm

There are 3 equations instead of just one because they are showing a 3 step process to get the final product, using each product as the reactant for the next problem!


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