How many grams of nh3 will be produced from a reaction between h2 and no if the partial pressures of h2 and no were 536 torr and 602 torr, respectively, and a temperature of 150.°c? assume the reaction volume is 25.0 l and the reaction yield is 93.5%.

There will be produced 5.398 grams of NH3

Explanation:

Step 1: Data given

Partial pressure H2 = 536 torr

Partial pressure of NO = 602 torr

Temperature = 150.0°C

Volume = 25.0 L

Reaction yield = 93.5 %

Step 2: The balanced equation

3H2 + 2NO → 2NH3 + O2

Step 3: Calculate moles of H2

p*V = n*R*T

⇒ with p = the pressure of H2 = 536 torr = 536/760 = 0.70526 atm

⇒ with V = the volume = 25.0 L

⇒ with n = the number of moles of H2 = TO BE DETERMINED

⇒ with R= the gas constant = 0.08206 L*atm/mol*K

⇒ with T =the temperature = 150.0 °C = 423 K

n= (0.70526*25.0)/(0.082061$423)

n = 0.508 moles H2

Step 4: Calculate moles NO

p*V = n*R*T

⇒ with p = the pressure of NO = 602 torr = 602/760 = 0.792165 atm

⇒ with V = the volume = 25.0 L

⇒ with n = the number of moles of NO = TO BE DETERMINED

⇒ with R= the gas constant = 0.08206 L*atm/mol*K

⇒ with T =the temperature = 150.0 °C = 423 K

n= (0.792165*25.0)/(0.08206*423)

n = 0.571 moles NO

Step 5: Calculate the limiting reactant

For 3 moles of H2 we need 2 moles of NO to produce 2 moles of NH3 and 1 mole of O2

H2 is the limiting reactant. It will be completely consumed (0.508 moles).

NO is in excess.

Step 6: Calculate moles NH3

For 3 moles of H2 we need 2 moles of NO to produce 2 moles of NH3

For 0.508 moles H2 we'll have 0.339 moles NH3 (This is when the reaction yield is 100%)

For reaction yield of 93.5 % the number of moles = 0.935 * 0.339 = 0.316965 moles NH3

Step 7: Calculate mass of NH3

Mass NH3 = moles NH3 * Molar mass NH3

Mass NH3 = 0.316965 moles NH3 * 17.03  5.398 grams

There will be produced 5.398 grams of NH3

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.