Solution (a)
Step 1: Write and balance the chemical equation.
4C₃H₆ + 6NO → 4C₃H₃N + 6H₂O + N₂
Step 2: Calculate the molar mass of each reactant.
For C₃H₆
molar mass = (12.01 g/mol × 3) + (1.008 g/mol × 6)
molar mass = 42.08 g/mol
For NO
molar mass = (14.01 g/mol × 1) + (16.00 g/mol × 1)
molar mass = 30.01 g/mol
Step 3: Determine the mole ratios needed.
For C₃H₆
mole ratio = 4 mol C₃H₆ : 4 mol C₃H₃N
For NO
mole ratio = 6 mol NO : 4 mol C₃H₃N
Step 4: Calculate the number of moles of C₃H₃N produced using the given masses of two reactants and their mole ratios.
Using C₃H₆
[tex]\text{moles of C₃H₃N = 126 g C₃H₆} × \frac{\text{1 mol C₃H₆}}{\text{42.08 g C₃H₆}} × \frac{\text{4 mol C₃H₃N}}{\text{4 mol C₃H₆}}[/tex]
[tex]\text{moles of C₃H₃N = 2.99 mol}[/tex]
Using NO
[tex]\text{moles of C₃H₃N = 175 g NO} × \frac{\text{1 mol NO}}{\text{30.01 g NO}} × \frac{\text{4 mol C₃H₃N}}{\text{6 mol NO}}[/tex]
[tex]\text{moles of C₃H₃N = 3.89 mol}[/tex]
Step 5: Determine the limiting and excess reactants.
Since C₃H₆ produces less number of moles of C₃H₃N
[tex]\boxed{\text{C₃H₆ is the limiting reactant.}}[/tex]
Since NO produces greater number of moles of C₃H₃N
[tex]\boxed{\text{NO is the excess reactant.}}[/tex]
Solution (b)
Step 1: Calculate the molar mass of C₃H₃N.
molar mass = (12.01 g/mol × 3) + (1.008 g/mol × 3) + (14.01 g/mol × 1)
molar mass = 53.06 g/mol
Step 2: Calculate the mass of C₃H₃N produced.
Since C₃H₆ is the limiting reactant
[tex]\text{mass of C₃H₃N = 2.99 mol C₃H₃N} × \frac{\text{53.06 g C₃H₃N}}{\text{1 mol C₃H₃N}}[/tex]
[tex]\boxed{\text{mass of C₃H₃N = 159 g}}[/tex]
Solution (c)
Step 1: Calculate the mass of NO consumed.
[tex]\text{mass of NO consumed = 2.99 mol C₃H₃N} × \frac{\text{6 mol NO}}{\text{4 mol C₃H₃N}} × \frac{\text{30.01 g NO}}{\text{1 mol NO}}[/tex]
[tex]\text{mass of NO consumed = 135 g}[/tex]
Step 2: Calculate the mass of NO left.
mass of NO left = initial mass of NO - mass of NO consumed
mass of NO left = 175 g - 135 g
[tex]\boxed{\text{mass of NO left = 40 g}}[/tex]
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