Chemistry II?

Consider the reaction in this experiment:

NaNO2 (aq) + H2NSO3H (s) → NaHSO4 (aq) + N2 (g) + H2O (l)

If you start with 1.627 g og H2NSO3H and an excess amount of NaNO2,

(A). How many moles of N2 will be produced?

(B). What volume will the N2 occupy at 25°C and 0.978 atm?


PS: can some1 show me steps how to solve this problem, thanks

Answer:
First of all, I'll tell you the steps:
1. You make sure that he equation is equilibrated
2. you calculate moles of H2NSO3H. For this, you have to find out the molecular mass for H2NSO3H:2+14+32+48+1= 97
1.627/97=0.017 moles H2NSO3H
3. To find out how many moles of N2 will be produced, you make a corresponding on the equation:
1 mole H2NSO3H........ mole N2
0.017 moles H2NSO3H........ moles N2
=> x=0.017 moles N2
4. To find out the volume, you must use the following equation:
pV=nRT V=(nRT)/p. You must use the value of pressure in N/m2: 1atm..1.013*10^5
0.978.y y=99071.4 N/m2
T=25+273= 298K, R= 8.314 J/mole*K
=> V=(0.017*8.314*298)/ 99071.4= 4.25*10^(-4) m3 =0.425 L

Whenever you have a substance that is in excess, you don't use it to calculate your substances!

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