Stoichiometry and gases, explain]

the following reaction takes place at stp. 4.40 l of nitrogen gas are formed. how many grams of ammonium nitrite decomposed?

2 nh4no2 (aq) > 2 n2 (g) + 4 h2o (l)

The mass of ammonium nitrite decomposed is 12.54 grams.

Explanation:

Pressure of the nitrogen gas = P = 1 atm

Volume of the nitrogen gas = V = 4.40 L

Temperature of the nitrogen gas = T = 273.15 K

Moles of nitrogen gas = n

PV = nRT (ideal gas equation)

According to reaction , 2 mole of nitrogen is obtained from 2 mole of ammonium nitrite.

Then 0.196 moles of nitrogen will  will be obtained from:

ammonium nitrite.

Mass of 0.196 moles of ammonium nitrite:

= 0.196 mol × 64 g/mol = 12.54 g

The mass of ammonium nitrite decomposed is 12.54 grams.

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