[zack]

sample of rock: 4.034g rock iron ore: Fe₂O₃ rock contains .231g of iron

a. what is the percent iron in the rock?

b. how many g of Fe₂O₃ are in the "rock" sample?

c. what is percent Fe₂O₃ in sample?

HELP PLEASE

a) percent is = molar mass of Fe/ molar mass of Fe₂O₃ times 100. = 55.85/159.7 x 100= 34.97%
b) .3497(4.034) =  1.4107g of Fe in rock sample
c) 100%  !!!!

[macman104]

You need to look at it differently.

The rock weighs 4.034 g and it contains 0.231 g of Iron.

You can calculate the grams of Fe₂O₃ in the rock from the grams of Iron. (Calculate the mass percent of Iron in Fe₂O₃, then you can use that with the grams of Iron).

From there you can calculate the percent of Fe₂O₃ in the rock sample.

[zack]

You need to look at it differently.

The rock weighs 4.034 g and it contains 0.231 g of Iron.

You can calculate the grams of Fe₂O₃ in the rock from the grams of Iron. (Calculate the mass percent of Iron in Fe₂O₃, then you can use that with the grams of Iron).

From there you can calculate the percent of Fe₂O₃ in the rock sample.

mass percent of iron is equall to .0231g/4.034g x 100=  5.73%

how can i use this with the grams of iron?

[ARGOS++]

Dear zak;

   a.)   Correct!    5.726% iron in the rock!
   b.)   0.231g iron corresponds to how much g Fe₂O₃?  Name it for the moment xy grams.
   c.)   How much % is xy g of 4.034 g?

I hope it will be of help to you.
Good Luck!
                    ARGOS⁺⁺

[zack]

Dear zak;

   a.)   Correct!    5.726% iron in the rock!
   b.)   0.231g iron corresponds to how much g Fe₂O₃?  Name it for the moment xy grams.
   c.)   How much % is xy g of 4.034 g?

I hope it will be of help to you.
Good Luck!
                    ARGOS⁺⁺

ok. 1 mole of Fe2O3 is equall to 159.7 and one mole of Fe is equall to 55.85. what do i need to start cancelling out to end with grams of fe2O3?

[ARGOS++]

[zack]

i did cross mulitiplication and came up with .6605g of fe203

[ARGOS++]

Dear zak;

You only forgot now that there are two Fe in one Fe₂O₃!

Good Luck!
                    ARGOS⁺⁺

[zack]

now i have .3303g of fe203
i don't want to do cross muiltiplicaton. how can i use dimensional analysis in solving this problem

[zack]

c) i have 8.19% of Fe2O3

[ARGOS++]

Dear zak;

NO!,  -  the halfe of 0.6605g is 0.3303g!

Make only the partial RxN:    2Fe    --->    1Fe₂O₃, and that makes 2 moles of Fe goes in 1 mole Fe₂O₃.

So:                                  2 * 55.85g  : 1 * 159.7g    =   0.231g  :  xy g
If you control then xy g = 0.3303g Fe₂O₃.

c.)  Is correct!

Good Luck!
                    ARGOS⁺⁺

[zack]

thanks alot

[ARGOS++]