If 0.120 g c3h6 and 0.10 g no are allowed to mix according to the following reaction, how many g n2 are produced? 4c3h6 + 6 no → 4c3h3n + 6h2o + n2

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

beta decay is the answer