Be sure to answer all parts. Ammonia is a principal nitrogen fertilizer. It is prepared by the reaction between hydrogen and nitrogen. 3H2(g) + N2(g) → 2NH3(g) In a particular reaction, 4.00 moles of NH3 were produced. How many moles of H2 and how many moles of N2 were reacted to produce this amount of NH3?

moles of H2

Mole ratio of H2 to NH3 is 3: 2

if 3: 2

? 4

= 4*3/2= 6

6moles of H2

moles of N2

mole ratio of N2 to NH3 is 1: 2

if 1 : 2

? 4

4*1/2 = 2

2 moles of N2

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

can you list the answers so i can answer