[Chemistry Questions][ASAP][+Rep]

[Lmp]

Need this done by the end of today. Amount of rep given will be dependent on how much you do.
Need work shown for most of these.

1. A gaseous compound contains 16.0 g of hydrogen and
96.0 g of carbon. If the molar mass of this compound is
28.06 g/mol, what is its molecular formula?

2. In a blast furnace, iron(III) oxide reacts with carbon
monoxide to produce iron and carbon dioxide.
(a) Identify the limiting reagent if 2.50 mol of iron(III)
oxide and 6.50 mol of carbon monoxide are available.
(b) Identify the limiting reagent if 200.0 g of iron(III) oxide
and 100.0 g of carbon monoxide are available.

3. When 8.40 g of zinc metal is placed in a solution in which
11.6 g of HCI(g) is dissolved, hydrogen gas and zinc
chloride are produced.
(a) Calculate the expected yield of hydrogen gas.
(b) If 0.19 g of hydrogen gas is produced, what is the
percentage yield?

4. A 27.5-mL sample of 0.112 mol/L CuSO4(aq) solution is
added to 45.0 mL of 0.088 mol/L Na2CO3(aq). A precipitate
is formed.
(a) Identify the limiting reagent in the reaction.
(b) Calculate the mass of CuCO3 that is produced in the
reaction.

5. Identify the two acid

[SmileYaDead]

Doesn't seem hard, I'll help you as much as I remember ;)

[Bettser]

1) C2H4 I believe. 12.01g/molx2 for carbon, leaves the rest to be hydrogen. (Im a dummy, forgot the 2)

2) I believe LR is just whichever one has the least moles :P

I would help with the rest but I need to get back to studying for my exam XD. Pm me later if you still need help with any.

[Kad]

I've heard Scarydark is good at it

[Ana]

(a) Identify the limiting reagent if 2.50 mol of iron(III)
oxide and 6.50 mol of carbon monoxide are available.
(b) Identify the limiting reagent if 200.0 g of iron(III) oxide
and 100.0 g of carbon monoxide are available.

2) Fe2O3 (s) + 3CO (g) -----> 2Fe (s) + 3CO2 (g)
a)For every mole of Fe2O3, you need 3 moles of CO2.
If you have 2.5 moles of Fe2O3, you need 2.5(3) moles of CO2, which is 7.5 moles. Since you don't have enough carbon monoxide, that is the limiting reagent.
b) Same math as above. If you have 200moles of Fe2O3, you need 200(3) moles of CO2, which is 600 moles. You don't have 600moles of CO2, so that means carbon monoxide is still the limiting reagent

[Brady]

Good luck getting help with this!
I don't remember anything from Chem., and I took it last year /:

[trix]

3. When 8.40 g of zinc metal is placed in a solution in which
11.6 g of HCI(g) is dissolved, hydrogen gas and zinc
chloride are produced.
(a) Calculate the expected yield of hydrogen gas.
(b) If 0.19 g of hydrogen gas is produced, what is the
percentage yield?

1)Zn + 2HCl ==> H^2 + ZnCl^2

2)Moles ratio = 1:1

3) 8.40/30 = 0.28 mols Zn
11.6/36.5= 0.32 mols HCl

0.28*1*2 = 0.56 mols H^2

mols * mr = mass

0.56 * 2 = 1.12g theoretical ( I think)

b) percentage yield = mass of actual/mass of theoretical * 100


0.19/1.12 * 100 = 16.95%

Sorry if I did it wrong.

[Ana]

Sorry, I'm doing these out of order xD

1)
16 g of H, 96 g of C, molecular mass: 28.06g
16g+96 = 112g (total amount of compound)

112g x (1mole/28.06g) = about 4 moles. This is the number of moles of the compound you have.

16g x (1mol/1g) = 16 moles. This is the total number of H moles you have.
16moles / 4moles = 4 moles. This is the number of moles of H you have per mole of compound.

96g x (1mol/12g) = 8 moles. This is the total number of C moles you have.
8moles / 4moles = 2 moles. This is the number of moles of C you have per mole of compound.

So in each mole of your compound you have 2 moles of C and 4 moles of H.

Compound: C2H4

---------- Post added at 05:43 PM ---------- Previous post was at 05:33 PM ----------

4) Na2CO3(aq) + CuSO4(aq) --> CuCO3(aq) + Na2SO4(aq)
Moles Ration 1:1
27.5mL x (.112mol/L) x (L/1000mL) = .00308 moles of Na2CO3
45mL x (.088mol/L) x (L/1000mL) = .00396 moles of Na2CO3
You have less Na2CO3, so that is the limiting factor.

Calculating the mass of CuCO3
63.55 + 12 + 3(16) = 123.55 g/mol of CuCO3

From the reaction, you would have .00308 moles of CuCO3
.00308moles x (123.55 g/mol) = .380 g