1a. The balanced equation is given below:
2NO + O2 β 2NO2
The coefficients are 2, 1, 2
1b. 755.32g of NO2
2a. The balanced equation is given below:
2C6H6 + 15O2 β 12CO2 + 6H2O
The coefficients are 2, 15, 12, 6
2b. 126.25g of CO2
Explanation:
1a. Step 1:
Equation for the reaction. This is given below:
NO + O2 β NO2
1a. Step 2:
Balancing the equation. This is illustrated below:
NO + O2 β NO2
There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by putting 2 in front of NO and 2 in front of NO2 as shown below:
2NO + O2 β 2NO2
The equation is balanced.
The coefficients are 2, 1, 2
1b. Step 1:
Determination of the limiting reactant. This is illustrated below:
2NO + O2 β 2NO2
From the balanced equation above, 2 moles of NO required 1 mole of O2.
Therefore, 16.42 moles of NO will require = 16.42/2 = 8.21 moles of O2.
From the calculations made above, there are leftover for O2 as 8.21 moles out of 14.47 moles reacted. Therefore, NO is the limiting reactant and O2 is the excess reactant.
1b. Step 2:
Determination of the maximum amount of NO2 produced. This is illustrated below:
2NO + O2 β 2NO2
From the balanced equation above, 2 moles of NO produced 2 moles of NO2.
Therefore, 16.42 moles of NO will also produce 16.42 moles of NO2.
1b. Step 3:
Conversion of 16.42 moles of NO2 to grams. This is illustrated below:
Molar Mass of NO2 = 14 + (2x16) = 14 + 32 = 46g/mol
Mole of NO2 = 16.42 moles
Mass of NO2 =?
Mass = number of mole x molar Mass
Mass of NO2 = 16.42 x 46
Mass of NO2 = 755.32g
Therefore, the maximum amount of NO2 produced is 755.32g
2a. Step 1:
The equation for the reaction.
C6H6 + O2 β CO2 + H2O
2a. Step 2:
Balancing the equation:
C6H6 + O2 β CO2 + H2O
There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by 6 in front of CO2 as shown below:
C6H6 + O2 β 6CO2 + H2O
There are 6 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 3 in front of H2O as shown below:
C6H6 + O2 β 6CO2 + 3H2O
There are a total of 15 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 15/2 in front of O2 as shown below:
C6H6 + 15/2O2 β 6CO2 + 3H2O
Multiply through by 2 to clear the fraction.
2C6H6 + 15O2 β 12CO2 + 6H2O
Now, the equation is balanced.
The coefficients are 2, 15, 12, 6
2b. Step 1:
Determination of the mass of C6H6 and O2 that reacted from the balanced equation. This is illustrated below:
2C6H6 + 15O2 β 12CO2 + 6H2O
Molar Mass of C6H6 = (12x6) + (6x1) = 72 + 6 = 78g/mol
Mass of C6H6 from the balanced equation = 2 x 78 = 156g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 15 x 32 = 480g
2b. Step 2:
Determination of the limiting reactant. This is illustrated below:
From the balanced equation above,
156g of C6H6 required 480g of O2.
Therefore, 37.3g of C6H6 will require = (37.3x480)/156 = 114.77g of O2.
From the calculations made above, there are leftover for O2 as 114.77g out of 126.1g reacted. Therefore, O2 is the excess reactant and C6H6 is the limiting reactant.
2b. Step 3:
Determination of mass of CO2 produced from the balanced equation. This is illustrated belowb
2C6H6 + 15O2 β 12CO2 + 6H2O
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 12 x 44 = 528g
2b. Step 4:
Determination of the mass of CO2 produced by reacting 37.3g of C6H6 and 126.1g O2. This is illustrated below:
From the balanced equation above,
156g of C6H6 produced 528g of CO2.
Therefore, 37.3g of C6H6 will produce = (37.3x528)/156 = 126.25g of CO2
