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

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the answer is d.) potassium