For the reaction ? NO + ? O2 → ? NO2 , what is the maximum amount of NO2 which could be formed from 16.42 mol of NO and 14.47 mol of O2? Answer in units of g. 003 1.0 points For the reaction ? C6H6 + ? O2 → ? CO2 + ? H2O 37.3 grams of C6H6 are allowed to react with 126.1 grams of O2. How much CO2 will be produced by this reaction? Answer in units of gram

1a. The balanced equation is given below:

2NO + O2 → 2NO2

The coefficients are 2, 1, 2

1b. 755.32g of NO2

2a. The balanced equation is given below:

2C6H6 + 15O2 → 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 → NO2

1a. Step 2:

Balancing the equation. This is illustrated below:

NO + O2 → 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 → 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 → 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 → 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 → CO2 + H2O

2a. Step 2:

Balancing the equation:

C6H6 + O2 → 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 → 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 → 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 → 6CO2 + 3H2O

Multiply through by 2 to clear the fraction.

2C6H6 + 15O2 → 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 → 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 → 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

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.