coletrain chemistry question

[coletrain]

This was a multiple choice question

5.00g of a mixture of magnesium and zinc was allowed to completely react with dilute sulfuric acid according to the equation;
M + H2SO4----> MSO4 + H2     (M is the mixture of zinc and magnesium)

If .250g of hydrogen gas was produced in the reaction, what would the percentage composition of the magnesium be?

the options are
a) 17.2%     
b) 82.8%
c) 37.6%
d) 62.4%

[appianway]

Did you leave something out? Because you don't have any information about how much magnesium or zinc specifically reacts...

[coletrain]

no thats all the information given

[appianway]

It doesn't seem to add up, because for ionic bonding to occur with the sulphate ion in the second part, what's denoted as M would need a charge of 2+, however, this would mean that charge wasn't conserved in the reaction...

[coletrain]

exactly the question is impossible. Ive literally spent the entire past week trying and gotten nowhere. I think it might have been a typo, cause no1 else was able to answer this.

[fady_22]

What is the answer supposed to be?

[hyperblade01]

This is definitely gonna be a long shot and I'm taking a wild guess here...


Alright so you know .250g of gas was produced. Find the mole amount which is .125mol

Based on the equation, 1 mol of M will produce 1 mol of and using this, you know that .125mol of the mixture is reacted

You also know that if Magnesium is x%, then Zinc must be (1-x)% of the mixture...the total mass must be 5.00g.

Let x = % composition of magnesium


You can now create an equation (which is based on mole*molar mass = mass):





Closest answer is D :S




EDIT: Failed right there

[coletrain]

the answer sheet says the answer is C) 37.6%

thanks for trying guys, but i'll just ask my teacher on monday. But im sure its prob a typo or something, i'll let you know

[stonecold]

that is weird.  the equation doesn't even tell you where the zinc ends up.

[coletrain]

I got another one, dont know whether it could be another typo seeing as its from the same sheet;

6.5L of oxygen is required to completely convert 1L of a gaseous hydrocarbon fuel to carbon dioxide and water. If the volumes were measured at the same temp. and pressure the hydrocarbon would be
a) CH4
b) C2H6
c) C4H10
d) C6H6

[stonecold]

is it C?

ill provide workings if so.

[coletrain]

is it C?

ill provide workings if so.

If you would be so kind

[stonecold]

Okay, so you you are working at the same pressure and temp, so I just chose to use S.T.P (standard temp and pressure) where 1 mol = 24.5 L

First convert everthing to mol.

N(O₂) = 6.5/24.5 = 0.265 mol

N(hydrocarbon) = 1/24.5 =0.041 mol

Now find the oxygen to hydrocarbon ratio.  i.e. divide n(O₂) by n(hydrocarbon)

0.265/0.041= 6.46

Now you just write balanced chemical equations for the combustion of each of the hydrocarbons in oxygen gas to form water and carbon dioxide.

This is the equation for C₄H₁₀

2C₄H₁₀ + 13O₂ --->  10H₂O + 8CO₂

Now divide n(O₂) by n(C₄H₁₀)

13/2= 6.5

This is very close to the oxygen to hydrocarbon ratio we found earlier, so we can deduce that the answer is C) C₄H₁₀

[coletrain]

Okay, so you you are working at the same pressure and temp, so I just chose to use S.T.P (standard temp and pressure) where 1 mol = 24.5 L

First convert everthing to mol.

N(O₂) = 6.5/24.5 = 0.265 mol

N(hydrocarbon) = 1/24.5 =0.041 mol

Now find the oxygen to hydrocarbon ratio.  i.e. divide n(O₂) by n(hydrocarbon)

0.265/0.041= 6.46

Now you just write balanced chemical equations for the combustions of each of the hydrocarbons in oxygen gas to form water and carbon dioxide.

This is the equation for C₄H₁₀

2C₄H₁₀ + 13O₂ --->  10H₂O + 8CO₂

Now divide n(O₂) by n(C₄H₁₀)

13/2= 6.5

This is very close to the oxygen to hydrocarbon ration we found earlier, so we can deduce that the answer is C) C₄H₁₀

Thanks a lot stonecold

[coletrain]

This is definitely gonna be a long shot and I'm taking a wild guess here...


Alright so you know .250g of gas was produced. Find the mole amount which is .125mol

Based on the equation, 1 mol of M will produce 1 mol of and using this, you know that .125mol of the mixture is reacted

You also know that if Magnesium is x%, then Zinc must be (1-x)% of the mixture...the total mass must be 5.00g.

Let x = % composition of magnesium


You can now create an equation (which is based on mole*molar mass = mass):





Closest answer is D :S




EDIT: Failed right there

Hey you got further than me bro