Ineed urgent grade at risk with my chemistry i beg anyone the image is posted and show work if you can

Hey! so I can help you with the beginning!

1.  The ratio of Fe to O2 is 4:3 so since they tell you that you have 56 moles of Fe, all you have to do is plug in the information. So 56/4 = 14 and 3 x 14 = 42. To conclude, the answer for number one is 42 moles of O2.

2.Same thing from #1 except the ratio is 4:2 (since we are now talking about Fe2O3) so you divide 56/4 and get 14 so you multipy 2x 14 which gives you 28. To conclude, the answr to numbner 2 is 28 moles of Fe2O3.

3. On this question it is asking for grams, not moles, so we have to look on our periodic table and look for the atomic mass of Fe. To start off the problem you are going to have to find the amount of moles of Fe needed to react with 27 moles of O2 so the ratio will be 4:3 so you need to divide 27/3 which gives you 9. Multiply 9x4 and that gives you 36 moles. DONT FORGET TO CONVERT TO GRAMS. The final answer will be 2,010.42 g of Fe after you multiply 36 x 55.845 g (Fe atomic weight.

I would help you more but I have school tomorrow as well... I help this helped though.

im only answering because the other person deserves brainliest

xplanation:

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o