How many grams of nitric acid will react completely with a block of iron metal that is 4.5 cm by 3.0 cm by 3.5 cm, if the density of iron is 7.87 g/mL? Show all steps of your calculation as well as the final answer. Fe + HNO3 → Fe(NO3)3 + H2
@Abhisar i need some assistance please
Can u do the question if i tell how many grams of iron is given ?
sorry i dont understand
i mean i don't understand the question you are asking
What part is giving u trouble in this (original question) question ?
i don't know how to solve the question. i do know that i first must balance the equation so i did, I came up with Fe2O3 + 6HNO3 yields 2Fe(NO3)3 + 3H2O. Then i don't know what to do after that.
All ryt let's seee !
First tell me the atomic masses of Fe, O, H & N
Fe: 55.845 ± 0.002 u O: 15.9994 ± 0.0004 u H:1.00794 ± 0.00001 u N: 14.0067 ± 0.0001 u like do you personal memorize the atomic mass or is it okay to look at a periodic table. Because i looked up the atomic masses.
wait i forgot where are the atomic masses located next to the element?
It's okay to look at the periodic table, but i personally remember atomic masses of common elements.
oh ok
How u balanced ur equation ?
Given equation is Fe + HNO3 → Fe(NO3)3 + H2 and ur balanced equation is Fe2O3 + 6HNO3 yields 2Fe(NO3)3 + 3H2O.
Getting it ?
i don't understand how to balance the equation.
Balanced equation will be 2Fe + 6HNO3 → 2Fe(NO3)3 + 3H2
Getting it ?
U can't change the reactant or product side for balancing the equation ! U changed Fe into Fe2O3
u can only add coefficients like 2Fe, 4Fe or whatever is needed
oh my bad!
Got it ?
yes i do.
Now from the balanced equation \(\large \sf 2 Fe + 6 HNO_3 = 2 Fe(NO_3)_3 + 3 H_2\) we can see that 2 moles (111.69 g) of Fe will react completely with 6 moles of nitric acid (378.07704 g).
sorry but where are you get the 2 moles of Fe?
So for 1 g of iron \(\huge\sf \frac{378.07}{111.69}\) = 3.38 g of Nitric acid will be required.
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Got it ?
oh oopss that was obvious, yes i got it.
Now we know that how much of nitric acid is required for each gram of iron. All we have to do now is to calculate the amount of iron in the block of iron.
agree ?
yes
So let's do it..Though it's not absolutely correct but we will have to assume that 1cm\(^3\) of iron = 1mL
So density of iron becomes 7.87 g/cm\(^3\)
Now what will be the volume of iron block ?
i don't really get it. So is it something like this: 7.87= 1 ml???
Don't see the above lines, just calculate the volume of the iron block by using this Volume = Length\(\times\)Breadth\(\times\)Height
Dimensions are given as 4.5 cm 3.0 cm 3.5 cm,
Volume=4.5\(\times\)3\(\times\)3.5=47.25
yes i was think that but i was sure!
Volume=4.5\(\times\)3\(\times\)3.5=47.25 cm\(^3\)
Given density of iron is 7.87 g/cm\(^3\) which means that 7.87 g of iron is present in 1 cm\(^3\). Can u tell me how much grams will be in 47.25cm\(^3\) ?
6.004???
7.87\(\times\)47.25 = ?
oh you have to mult.? i divide oops so it is 371.8575
Now remember what we calculated earlier ? \(\color{blue}{\text{Originally Posted by}}\) @Abhisar So for 1 g of iron \(\huge\sf \frac{378.07}{111.69}\) = 3.38 g of Nitric acid will be required. \(\color{blue}{\text{End of Quote}}\)
yes
Now for 371.8575g of iron how much nitric acid will be required ?
should I multiply 371.8575 X 3.89?
obvsly !
sorry i am not the brightest person, i got 1,446,52568
\(\color{blue}{\text{Originally Posted by}}\) @superhelp101 should I multiply 371.8575 X \(\cancel{3.89}\) 3.38 \(\color{blue}{\text{End of Quote}}\)
Got it ?
eh. then the answer is 1256.87835. wow i can't believe how one decimal point makes such a big difference to the answer.
Yep ! Copy carefully :)
and u were wrong ! u r the brightest person around here, even brighter than me !
What? Yayy. You made my day even though even though it is like 1:00am. Lol how come on my screen it shows that i only have 64 points?? that that is weird?
\(\huge\color{green}{\text{¯\_(ツ)_/¯}}\)
did you edit the snapshot?? to make it look like i had 99 points?
Ahahahahhahaha that is sooooooo funny
^_^
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