Chemistry Question ? plz I need help.?

Question

1) 5.34g of a salt of formula M2SO4 (Where M is a metal) were dissolved in water. The sulphate ion was precipitated by adding excess barium chloride solution when 4.66g of barium sulphate (BaSO4) were obtained .

(a)How many moles of sulphate ion were precipitaed as barium sulphate?

(b)How many moles of M2SO4 were in the solution?

(c)What is the formula mass of M2SO4?

(d)What is the relative atomic mass of M?

(e)Use a table of relative atomic masses to identify M.

(2)(a)Explain what is meant by "hydrogen peroxide has the molecular formula H2O2'

(b)2H2O2------>2H2O2 + O2
What mass of oxygen can be formed from 17g of H2O?

(c)What mass of oxygen can be formed from 17g of H2O2 if 1 g of maganese dioxide catalyst is used?

(3)What mass of solid is precipitated when excess dilute sulphuric acid is added to 100cm^3 of 2 M barium chloride solution?

Explain plz.

Answer 1

These aren't questions from hell, they're perfectly reasonable stoichiometry questions. I'm not going to solve them for you, you should be able to do that. Let me remind you of some things, though. First, you shouldn't have any problem converting a mass of something to moles using the molecular weight, right? You also need to think about balanced equations. Once you have those things, you just have to think about how you can get from what you know to what you need to figure out.

So, for the first question, you know the mass of BaSO4 that you produced. So, you can calculate its molar mass and convert the 4.66 grams to moles of BaSO4. Since you know that 1 mol of BaSO4 contains one mole of SO4(2-) ions, you know how many moles of sulfate ions came from the original stuff that was dissolved. That has to be the same as the number of moles that is contained in 5.34 g of the M2SO4 salt. So, if you divide 5.34 by the moles of sulfate ions, you have the molar mass of the M2SO4 salt. If you then take the molar mass of M2SO4 and subtract the mass of a sulfate ion from it, you're left with 2 times the atomic mass of the metal M.

In your question 2b, you have the formula wrong. THe products of the reaction are oxygen gas (O2) and water H2O. Assuming that it should be "What mass of oxygen can be formed from 17 g of H2O2?", here again, convert the mass of H2O2 into moles using the molecular weight of H2O2. Then from your balanced equation it takes 2 mol H2O2 to give you 1 mol of O2. Once you figureout how many moles of O2 you can get, you can convert that to grams. For (c), adding a catalyst makes the reaction happen more quickly, but can't change the amount of the product you form. So, 2b and 2c have the same answer.

I'll let you work on 3. Convert the volume to L and use the concentration to get to moles of barium chloride. Then, use a balanced equation to figure out how many moles of barium sulfate you can form, and then convert that to moles.

The process is basically the same in all of these problems.

Answer 2

(a). Molar mass (Mr) of BaSO4 equals 233, so the 4.66 g of BaSO4 are:

n = m/Mr = 4.66/233 = 0.02 mol BaSO4. That is the answer to your first question.

(b). Now, from the chemical equation:

M2SO4 + BaCl2 -> 2MCl + BaSO4

we see that 1 mol of M2SO4 produces 1 mol of BaSO4, when it reacts fully (that's our case because BaCl2 is in excess). So there were 0.02 moles of M2SO4 in the solution.

(c). Formula mass (or molar mass) of M2SO4 equals:

Mr = m/n = 5.34/0.02 = 267.

(d). Mr = 267, but:

Mr = 2Ar(M) + Ar(S) + 4*Ar(O) = 2Ar(M) + 32 + 4*16 = 2Ar(M) + 96 (where Ar are the relative atomic masses)

Solving the equation: 2Ar(M) + 96 = 267, we get:

Ar(M) = 85.5

(e). From the tables we get that M is the element Rb (Rubidium).

2.

(a) That means that the molecule of H2O2 consists of 2 atoms of hydrogen and tow atoms of oxygen (H-O-O-H).

(b). The molar mass of H2O2 is 34, so the 17 g of H2O2 are:

n = m/Mr = 17/42 = 0.5 mol.

Because 2 moles of H2O2 give 1 mol of oxygen the 0.5 mol will give 0.5/2 = 0.25 mol O2.

The molar mass of O2 is 32 so the 0.25 mol are:

m = n*Mr = 0.25*32 = 8 g of O2.

(c). Again 8 g of O2. The catalyst (MnO2) makes the reaction to go faster, but the same amount of O2 is produced.

(d). The solution contains:

n = C*V = 0.1 L * 2 M = 0.2 mol of BaCl2.

From the reaction:

BaCl2 + H2SO4 -> BaSO4 + 2HCl

we see that 1 mol of BaCl2 give 1 mol of BaSO4. Because H2SO4 is in excess, BaCl2 reacts fully, so we get 0.2 mol of BaSO4, with a molar mass of 233 and a mass of:

m = n*Mr = 0.2*233 = 46.6 g.

That's it, I thing i deserve the 10 points, thank's.

Answer 3

a) 4.66 grams / 233 barium sulphate = 0.02 moles

b) Same as A

c) 5.34 grams / 0.02 moles = 267

d) 267 - 96 (SO4) / 2 = 75.5

2a) 2 atoms of hydrogen and 2 atoms of oxygen

b) Assuming H2O2 one would get 16.5 grams of H2O2

Answer 4

i help you with the 1st part i'm too lazy to type the rest :-)
4.66g of barium sulphate (BaSO4) were obtained .
(a)How many moles of sulphate ion were precipitaed as barium sulphate?
calc the no of moles of BaSO4 by dividing the mass obtained by the rel mol mass(Mr) of BaSO4. there is 1 mol of SO4 ions in 1 mol of BaSO4. so the no of mol of SO4 ions = no of mol of BaSO4 obtained

Answer 5

Hey, this won't take me several hours. I took AP Chem last year as a senior in H.S. and this was a standard problem-takes 15 mins. I might have time to do some. I'll answer back in a bit.

Answer 6

I'm a chem teacher and this one would take a couple of hours to solve....you must have the professor from hell....good luck