3. ammonia, nh3, is a typical ingredient in household cleaners. it is produced through a combination reaction involving n2(g) and h2(g). if 24.0 mol of h2(g) react with excess n2(g), how many moles of ammonia are produced?

16 moles of Ammonia are produced

Explanation:

First thing's first, we have to write out the balanced chemical equation for the production of ammonia.

We know that;

Nitrogen + Hydrogen --> Ammonia

N2 + 3H2 --> 2NH3

From the equation, we can tell that 1 mol of N2 reacts with 3 mol of H2 to produce 2 mol of NH3.

Since N2 is in excess, it cannot be a limiting reactant. So the mole relationship is reduced to;

3 mol of H2 produces 2 mol of NH3

24 mol of H2 produces x mol of NH3

3 = 2

24 = x

Upon cross multiplication and solving for x;

x = ( 24 * 2) / 3

x = 16 mol

yes

explanation:

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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