I have another question c: 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 @ganeshie8
Yay (\^-^/) chemistry!
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`4.5 cm by 3.0 cm by 3.5 cm` start by finding the total mass of the iron block
Volume of block = ?
Mass = Volume x density = ?
Do I just multiply them together?
yess
47.25
yes 47.25 cm^3 us the volume. what about mass ?
So 47.25/7.87=6.0
you need to multiply : Mass = Volume * density
Ohh that's a big difference .-. 371.8575
Yes! so we have a 371.8575 g of iron metal ready for reaction with nitric acid
lets see how much nitric acid it takes
convert 371.8575 g to moles
http://www.chemicool.com/elements/iron.html 1 mol of Fe weighs 55.85 g so how many mols are present in 371.8575 g of Fe ?
Okay so 371.8575/6.02x10^23
nope, number of mols present in 371.8575 g of Fe = 371.8575 / 55.85
= 6.66 mol
that means we need 6.66 mol of nitric acid for the reaction as well
Find the molar weight of HNO3, then multiply by 6.66
H : 1 N : 14 O3 : 16*3 = 48 ---------------- HNO3 : 1+14+48 = 63 g
1 mol of HNO3 weight 63g
6.66 mol would weigh how much ?
You said you don't know much about chem, you know enough to make me have a headache XD and it's 419.46
thats it ! lol this is not chemistry, this is just math.
.-. chem is basically all math... Ad I suck at math.... That's why I had to retake both precancerous honors and chemistry honors .-. Because I can't understand them
XD oh my god it autocorrected precac to precancerous
haha idk what autocorrect has in mind lol... you're good in math too ! but im not seeing you much math recently hmm
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