Ammonia gas and oxygen gas react to form water vapor and nitrogen monoxide gas. what volume of nitrogen monoxide would be produced by this reaction if of oxygen were consumed?

The question is incomplete.

The probable question is

Ammonia gas and oxygen gas react to form water vapor and nitrogen monoxide gas. what volume of nitrogen monoxide would be produced by this reaction if 6.3 L of oxygen were consumed?

-

5.04 L

Explanation: -

The balanced chemical equation for this process is

4NH₃ + 5O₂ → 4NO + 6H₂O

Using gay lussac law of combining volumes from the balanced chemical equation we see

5 volumes of O₂ react to give 4 volumes of NO

6.3 L of oxygen will give x 6.3 L

= 5.04 L of NO

The volume of NO produced by 6.3 L of oxygen is 5.04 L.

Explanation:

The question states the formation of water vapors and nitrogen monoxide in the reaction of Ammonia with Oxygen.

If the amount of oxygen consumed  = 6.3 L, the volume of NO will be?

This implies that,

5 moles of oxygen produces 4 moles of NO.

So, 6.3 L of Oxygen produces = 6.3 L

NO produced by 6.3 L of oxygen = 5.04 L.

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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

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