Emergency help needed for chem just one question. (1 Viewer)

[Shoom]

Caclualte the mass of magnesium oxide formed when 10g of magnesium burns in excess oxygen.

Now the balanced equation is 2Mg+02 ----- 2MgO

Now n= 10/24.31 = 0411

m= 0.411 x ( 24.31+24.31+32) = 33.16

Why does my book divide 0.411 by 2 if the mol ratio is 1:1.


Last question

Calculate the mass of nitrogen in 28.02g of ammonium nitrate NH4NO3

n= 28.02/80.05 = 0.35000

Caclulate the number of moles of N atmos why is their 2 mols of Nitrogen shouldnt it just be 1?

 

[tommykins]

Caclualte the mass of magnesium oxide formed when 10g of magnesium burns in excess oxygen.

Now the balanced equation is 2Mg+02 ----- 2MgO

Now n= 10/24.31 = 0411

m= 0.411 x ( 24.31+24.31+32) = 33.16

Why does my book divide 0.411 by 2 if the mol ratio is 1:1.

That's where you went wrong.

Last question

Calculate the mass of nitrogen in 28.02g of ammonium nitrate NH4NO3

n= 28.02/80.05 = 0.35000

Caclulate the number of moles of N atmos why is their 2 mols of Nitrogen shouldnt it just be 1?

Theres one in NH4 and another in NO3, making it 2 mols of nitrogen (ie. 0.7 moles from your working).

 

[Shoom]

So what did I do wrong in the first question? OH is it because 2:2 is the same as 1:1 so I multiply by it by 24.31+16 only correct?

 

[tommykins]

The mass of MgO is only 24.31+16, 2MgO means you have 2 moles of MgO.

But yes basically whta you jsut said.

 

[Shoom]

But if I were to balance an equation and I had to put a 2 infront of H2O does that mean I have 4 hydrogen and 2 oxygen?

If their was a 3 infront of MgO it would be 3/2 x 0.411 x ( 24.31+16)? or 3/2 x0.411 x ( 24.31 x3 + 16x3)?

 

[tommykins]

Yes.

 

[Shoom]

and the only reason the Molar Mass of MgO is 3 x 24.31 + 3 x16 is because the ratios are uneven if they were even e.g 2:2 3:3 then its just (24.31+16)?

thanks for help

 

[tommykins]

the molar mass of MgO is ALWAYS 24.31+16, the number in front (ie. 2MgO) is just how many moles of MgO we have.

 

[Shoom]

so if the equation was kept the same but a 3 was infront of the mgo what would the answer be show working thanks

 

[tommykins]

Now the balanced equation is 2Mg+02 ----- 3MgO

Now n= 10/24.31 = 0.411

We have a ratio of 2:3
Thus n(MgO) = 0.411/2 * 3 = 0.6165 moles

m(MgO) = 0.6165*(24.31+16) = 24.85grams

 

[Shoom]

k so even if it's 60mgo the molar mass is still 24.31+16 so basically I should ignore the number infront of the product when calculating molar mass

 

[tommykins]

yes.

 

[ratcher0071]

Now the balanced equation is 2Mg+02 ----- 3MgO

How is that:
You got 2Mg and 20 on the reactant side and
you got 3Mg and 3O on the product side

 

[kaz1]

How is that:
You got 2Mg and 20 on the reactant side and
you got 3Mg and 3O on the product side

I'm pretty sure it's a typo.

 

[tommykins]

Read the post before mine, it was providing a random example in the effect of 3MgO instead of 2MgO.

 

[ratcher0071]

Read the post before mine, it was providing a random example in the effect of 3MgO instead of 2MgO.

ok sorry