Answer:
1] Given the equation: 2 NH3 (g) + 3 Cl2(g) ---> N2(g) + 6 HCl(g)
a. How many milliliters of nitrogen can be made from 13 L of chlorine and 10.0 L of ammonia gas at STP?
10.0 L NH3 X
1 L N2
2 L NH3 = 5.00 L N2
13 L Cl2 X
1 L N2
3 L Cl2
= 4.3 L N2
Answer ________4.3 x 103 mL N2_________
b. How many grams of chlorine must react to produce 16 L of nitrogen gas at 1.2 atm and 23oC?
nN2 = (1.2atm) x (16L)
(0.0821
L-atm
mol-K) x (273+23)K
= 0.790 mol N2
0.790 mol N2 X
3mol Cl2
1mol N2
X 70.1g Cl2
1mol = 1.66 x 102 L Cl2
Step-by-step explanation: