[EMTMATT]

I am stuck on only one chemistry question....

This seems to be a difficult topic for me...The questions are...

Reacting gaseous ammonia (NH3) with carbon dioxide produces solid urea (NH2)2CO and water.  Calculate how many grams of urea would be produced from 55.7 grams of ammonia.

2NH3 + CO2 -----4H2O + 2(NH2)2CO

I balanced the equation, but don't know how to approach it.

[Sev]

How many moles NH₃ in 55.7g?  How many moles of urea will this react for?

edit: your eqn is not balanced either.

[EMTMATT]

Sorry...I guess I will need help starting with the basics of balancing the equation.


___NH3 + _____CO2 ----- _____H2O + ______(NH2)2CO

Above is the non-balanced equation.  Can you show me step-by-step on how to solve the problem on my original post?  I am very confused.  I am sorry to bother you.

[Sev]

I am sorry to bother you.

No bother at all. 

The rxn is: 2NH₃ + CO₂ → H₂O + (NH₂)₂CO + H₂O

You start with 55.7g NH₃.  First calculate the number of moles this is (molar mass of NH₃ is 17.04g.mol^-1).
n = m/M

Use this value to calculate the number of moles of urea formed (use mole ratios from rxn eqn). 
Now find mass of urea, given that molar mass is 60.07g.mol^-1
m = nM

Hope that helps.

[EMTMATT]

that helps a lot, thank you.

I have one last question, then I will bother you again tonight.

The reaction N2 +3H2 ---- 2NH3 produces ammonia.  If three moles of nitrogen gas reacts with eight moles of hydrogen gas, what is the limiting reactant and how many moles of ammonia are produced?

Thank You

[Sev]

The reaction N2 +3H2 ---- 2NH3 produces ammonia.  If three moles of nitrogen gas reacts with eight moles of hydrogen gas, what is the limiting reactant and how many moles of ammonia are produced?

H₂ is limiting reagent (since N₂ is in excess).
If 3 mol H₂ reacts for 2 mol NH₃, 8 mol H₂ will react for 8*(2/3) = 5.33 mol NH₃.

[Borek]

Read 2^nd paragraph to learn how to read balanced reaction equation:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations