excess reactant...

[Moe_slow]

ok...i know how to find the limiting reactant but am lost as in how to find out how much is left over...

Example:

N2H4 + 7H2O2 ---> 2HNO3 + 8H2O

a) 120g of N2H2 reacts with an equal mass of H2O2. Which is the limiting reactant.

i got is done:
N2H2= M/mm=120g / 32.1 g/m = 3.74 moles
H2O2= M/mm=120g / 34.02 g/m = 3.5 moles

hence H2O2 is the limiting reactant.

b) What mass of HN03 is expected?

2/x = 7/3.5
7x=7
x=1 mole

HN03= 63.02 g/mole X 1 mole = 63.0 g

C) What mass, in grams, of the excess reactant remain at the end of the reaction?

this is where im lost...some help would be nice...

[mrjeffy321]

In part a), to find the limiting reactant you found the number of moles present of each of the reacts.

N2H2= M/mm=120g / 32.1 g/m = 3.74 moles
H2O2= M/mm=120g / 34.02 g/m = 3.5 moles

You know from the chemical equation that the N2H4 reacts in a 1:7 ratio with H2O2,
N2H4 + 7H2O2 --->
1 mole of N2H4 : 7 moles H2O2

If you know that H2O2 is the limiting reactant, you know that after the reaction, none remains. All the H2O2 needed to react with the excess reactant, N2H4 in its propper molar ratios. For every 3.5 moles of H2O2 reacted, .5 moles of N2H4 are needed (3.5 / 7). Thus, after the reaction, you have .5 moles less of N2H4 than you did when you started, for a remaining 3.24 moles.
Converting this number of moles to grams is simply a matter of multiplying by the molar mass.

[Moe_slow]

thanks i got it...just a bit confused...now its clear...thanks again :wink:

[Moe_slow]

Can I get some assistance with concentration problems?

1. What mass of formaldehyde, CH20 is contained in 250 mL of 8.0 mol/L solution of formaldehyde.

I went about this with first getting the molar mass of formaldehyde; which is 30 g/mol. Then since 8.0 mole/L I divided it by 4 to get 2 mol/250 mL.

Mas= MM x moles
=30 g/mol x 2.0
=60 g

Is this correct?

2. Calculate the concentration of sucrose solution when 5.0 g of sucrose, C^12 H^22 O^11, is added to water make 50.0 mL of solution.

I did this in a mass/volume percent formula.

Mass/volume= (mass of solute(g)/Volume of solution) X100%
=(5.0g /50.0mL)X100%
=10g/ml

Correct?

3. What volume of a 6.0 M solution of potassium nitrate would contain 3.0g of solute.

I am stuck hear a bit. Do I need to use molar mass of potassium to get its mass and then use a mass/mass formula to get the answer?

4. Stock solutions of sulphuric acid are 96.0 % H2SO4 by mass. These stock solutions have a density of 1.84 g/mL. Calculate the concentration of these stock solutions of sulphuric acid in mol/L.


I need mad help to start these off...am a bit thrown off...dont know where to start...


5. Stock solutions of hydrochloric acid have a concentration of 11.66 mol/L and a density of 1.18
g/mL. Calculate the concentration in units of mass/mass percent of these stock solutions.

same as 4...need a place to start...

[Borek]

Check out this lecture on percentage to molarity conversion (http://www.chembuddy.com/?left=concentration&right=percentage-to-molarity).

Best,
Borek
--
Chemical calculator (http://www.chembuddy.com/?left=chemical_calculators)s at www.chembuddy.com (http://www.chembuddy.com)
equation balancer and stoichiometry calculator (http://www.chembuddy.com/?left=EBAS&right=equation-balancing-stoichiometry)
www.pH-meter.info (http://www.ph-meter.info)

[reincarnated_soul]

Can I get some assistance with concentration problems?

1. What mass of formaldehyde, CH20 is contained in 250 mL of 8.0 mol/L solution of formaldehyde.

I went about this with first getting the molar mass of formaldehyde; which is 30 g/mol. Then since 8.0 mole/L I divided it by 4 to get 2 mol/250 mL.

Mas= MM x moles
=30 g/mol x 2.0
=60 g

Is this correct?

2. Calculate the concentration of sucrose solution when 5.0 g of sucrose, C^12 H^22 O^11, is added to water make 50.0 mL of solution.

I did this in a mass/volume percent formula.

Mass/volume= (mass of solute(g)/Volume of solution) X100%
=(5.0g /50.0mL)X100%
=10g/ml

Correct?

3. What volume of a 6.0 M solution of potassium nitrate would contain 3.0g of solute.

I am stuck hear a bit. Do I need to use molar mass of potassium to get its mass and then use a mass/mass formula to get the answer?

4. Stock solutions of sulphuric acid are 96.0 % H2SO4 by mass. These stock solutions have a density of 1.84 g/mL. Calculate the concentration of these stock solutions of sulphuric acid in mol/L.


I need mad help to start these off...am a bit thrown off...dont know where to start...


5. Stock solutions of hydrochloric acid have a concentration of 11.66 mol/L and a density of 1.18
g/mL. Calculate the concentration in units of mass/mass percent of these stock solutions.

same as 4...need a place to start...


For concentration problems u must know the almighty formulae:
No. of moles= Concentration x Volume (when the volume is given in L or dm^3)
No. of moles=Concentration x Volume/1000 (when the volume is in mL or cm^3)
And also, No. of moles=Mass/Molar Mass
Then ur all set!

1. First of all ur solution is not correct.:surprised

No. of moles= C x V/1000
=8 x 250/1000
=2 moles.

Molar mass of CH20 = 12 + (1 x 20)= 32 g/mol

No. of moles=Mass/Molar Mass
Mass=No. of moles x Molar Mass
=2 x 32=64g

2. I am really surprised at the way u approach to solve these problems......I really wonder how u come up with the concepts u stated....read a couple of good books in chemistry..........

Molar Mass of sucrose= (12 x 12) + (1 x 22) + (16 x 11)= 342g/mol

No. of moles of sucrose=5/342=0.0146 mol

No. of moles= C x V/1000
0.0146= C x 50/1000
C=0.292mol/L

3. The molecular formula for potassium nitrate is KNO3.
Molar mass of KNO3= 39 + 14 + (16 x 3) = 101g/mol

No. of moles of KNO3=3/101=0.0297 mol

No. of moles= C x V/1000
0.0297 = 6 x V/1000
V=4.95 mL

Well i gotta go to sleep now:zzz: , the rest is for u to find out..........
***B***

[rawr.panda.55]

If you know that H2O2 is the limiting reactant, you know that after the reaction, none remains. All the H2O2 needed to react with the excess reactant, N2H4 in its propper molar ratios. For every 3.5 moles of H2O2 reacted, .5 moles of N2H4 are needed (3.5 / 7). Thus, after the reaction, you have .5 moles less of N2H4 than you did when you started, for a remaining 3.24 moles.
Converting this number of moles to grams is simply a matter of multiplying by the molar mass.

Is it possible for you to show this in a calculation formate, because em not getting the meaning behind the words :| PLEASE AND THANKS! <3