The reaction forms 0.016 g N_2.
We have the masses of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. Gather all the information in one place with molar masses above the formulas and everything else below them.
M_r: __42.08 ___30.01 ___________________28.01
_____4C_3H_6 + 6NO → 4C_3H_3N + 6H_2O + N_2
Mass/g: 0.120 ___0.10
Step 2. Calculate the moles of each reactant
Moles of C_3H_6 = 120 mg C_3H_6 × (1 mmol C_3H_6 /42.08 mg C_3H_6 ) = 2.852 mmol C_3H_6
Moles of NO = 100 mg NO × (1 mmol NO/30.01 mg NO) = 3.33 mmol NO
Step 3. Identify the limiting reactant
Calculate the moles of N_2 we can obtain from each reactant.
From C_3H_6: Moles of N_2
= 2.852 mmol C_3H_6 × (1 mmol N_2/4 mmol C_3H_6 ) = 0.7129 mmol N_2
From NO: Moles of N_2 = 3.33 mmol NO × (1 mmol N_2/6 mmol NO)
= 0.555 mmol N_2
NO is the limiting reactant because it gives the smaller amount of N_2.
Step 4. Calculate the mass of N_2.
Mass = 0.555 mmol N_2 × 28.01 mg N_2/1 mmol N_2 = 16 mg N_2
= 0.016 g N_2
Taking into account the reaction stoichiometry, 0.0155 grams of N₂ are produced when 0.120 grams of C₃H₆ reacts with 0.10 grams of NO.
In first place, the balanced reaction is:
4 C₃H₆ + 6 NO → 4 C₃H₃N + 6 H₂O + N₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The molar mass of the compounds is:
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction and a simple rule of three as follows: if by stoichiometry 168 grams of C₃H₆ reacts with 180 grams of NO, if 0.120 grams of C₃H₆ react how much mass of NO will be needed?
[tex]mass of NO=\frac{0.120 grams of C_{3} H_{6} x180 grams of NO}{168 grams of C_{3} H_{6}}[/tex]
mass of NO= 0.128 grams
But 0.128 grams of NO are not available, 0.10 grams are available. Since you have less mass than you need to react with 0.120 grams of C₃H₆, NO will be the limiting reagent.
Considering the reaction stoichiometry and the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 180 grams of NO form 28 grams of N₂, 0.10 grams of NO form how much mass of N₂?
[tex]mass of N_{2} =\frac{0.10 grams of NOx28 grams of N_{2} }{180 grams of NO}[/tex]
mass of N₂= 0.0155 grams
Finally, 0.0155 grams of N₂ are produced when 0.120 grams of C₃H₆ reacts with 0.10 grams of NO.
Learn more about the reaction stoichiometry:
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