iron (II) sulfide reacts with hydrochloric acid to produce iron (II) chloride and hydrogen sulfide gas?

Question

how many grams of acid are needed to react with 75.0 grams of iron (II) sulfide ore which contains 30.0% inert material?

please provide the formula for me

Answer 1

First generate the balanced equation for the reaction. Note; we need 2 HCl's to make everything come out OK:

FeS + 2 HCl ---> FeCl2 + H2S

Now, correct for actual composition of iron(II) sulfide in 70% pure ore:

75g x [100 - 30]% = 52.5g of FeS present in the sample:

Determine no of moles of FeS present:

52.5g/[55.85 + 32.06]g/mole = 0.597mole

Now, from the ratios of the balanced equation and the number of moles of iron sulfide present you can calculate moles of HCl reqired.

0.597mole x (2 mole HCl/mole FeS) = 1.194 moles HCl.

To get grams of HCl needed, multiply moles HCl by grams HCl:

1.194moles x (35.45 + 1.01) = 43.5g HCl

You could put all the steps together as:

(75g x 0.70%) x (2/1) x (35.45 + 1.01)g/mole/(55.85 + 32.06)g/mole = 43.5g HCl

Remember, this is the weight of HCl as a pure gas that would be required; the hydrochloric acid you have in the lab is HCl dissolved in water so the weight of HCl solution required would depend on the concentration.

Answer 2

Iron Sulfide And Hydrochloric Acid

Answer 3

Just giving you the answer doesn't help you learn anything. You need to understand the process of how to do these types of problems in chemistry because they are VERY important and don't go away as you progress through school. First, translate the words into the atomic symbol for each: HCl(aq) + FeS(s) ---> FeCl2(aq) + H2S(g) To make sure it's balanced, just count each atom on each side Hydrogens (H) - One on left, two on right Chloride (Cl) - One on left, two on right Iron (Fe) - One on left, one on right Sulfur (s) - One on left, one on right So, you need one more Hydrogen on the left and one more Chloride on the left. The easiest way to do this is to put a 2 in front of HCl. Remember that the 2 in front means that you multiply each by that number. 2 HCl(aq) + FeS(s) ---> FeCl2(aq) + H2S(g) Now there are 2 hydrogens, 2 chlorides, 1 iron, and 1 sulfur on the left and the same on the right. Hope this was helpful.

Answer 4

FeS + 2HCl --> FeCl2 + H2S

If the ore contains 30.0% inert material, it contains 100 - 30 = 70% reactive material.

g, FeS = 75.0 g ore x 0.700 = 52.5g FeS

Moles FeS = 52.5g FeS x 1 mol FeS/ 87.95g FeS
= 0.597 mol FeS

Moles HCl = 0.597 mol FeS x 2mol HCl/1mol FeS
= 1.194 mol HCl

g, HCl = 1.194 mol HCl x 36.45g/mol = 43.5 g HCl

Ans: 43.5 g HCl to 3 sf