Question

what are the values of 2C6H6 + 15O2 → mCO2 + nH2O in this reaction?

Answer 1

2C6H6(l) + 15O2(g) --> 12CO2(g) + 6H2O(l) ... balanced equation

2C6H6(l) + 15O2(g) --> 12CO2(g) + 6H2O(l) ... balanced equationThe ∆H for the reaction, as written, is -6542 kJ. This represents the heat generated from burning 2 moles of C6H6(l) and generating 12 moles of CO2(g), etc.

2C6H6(l) + 15O2(g) --> 12CO2(g) + 6H2O(l) ... balanced equationThe ∆H for the reaction, as written, is -6542 kJ. This represents the heat generated from burning 2 moles of C6H6(l) and generating 12 moles of CO2(g), etc.From the mass of C6H6 given, we can calculate moles of C6H6 combusted. From that value, we can then find the heat that would be generated. After finding the heat generated, we can find the change in temperature of the water.

2C6H6(l) + 15O2(g) --> 12CO2(g) + 6H2O(l) ... balanced equationThe ∆H for the reaction, as written, is -6542 kJ. This represents the heat generated from burning 2 moles of C6H6(l) and generating 12 moles of CO2(g), etc.From the mass of C6H6 given, we can calculate moles of C6H6 combusted. From that value, we can then find the heat that would be generated. After finding the heat generated, we can find the change in temperature of the water.5.300 g C6H6 x 1 mol/78.11 g = 0.06785 moles

2C6H6(l) + 15O2(g) --> 12CO2(g) + 6H2O(l) ... balanced equationThe ∆H for the reaction, as written, is -6542 kJ. This represents the heat generated from burning 2 moles of C6H6(l) and generating 12 moles of CO2(g), etc.From the mass of C6H6 given, we can calculate moles of C6H6 combusted. From that value, we can then find the heat that would be generated. After finding the heat generated, we can find the change in temperature of the water.5.300 g C6H6 x 1 mol/78.11 g = 0.06785 molesheat generated = 0.06785 moles x 6542 kJ/2 moles = 221.9 kJ

2C6H6(l) + 15O2(g) --> 12CO2(g) + 6H2O(l) ... balanced equationThe ∆H for the reaction, as written, is -6542 kJ. This represents the heat generated from burning 2 moles of C6H6(l) and generating 12 moles of CO2(g), etc.From the mass of C6H6 given, we can calculate moles of C6H6 combusted. From that value, we can then find the heat that would be generated. After finding the heat generated, we can find the change in temperature of the water.5.300 g C6H6 x 1 mol/78.11 g = 0.06785 molesheat generated = 0.06785 moles x 6542 kJ/2 moles = 221.9 kJq = mC∆T

2C6H6(l) + 15O2(g) --> 12CO2(g) + 6H2O(l) ... balanced equationThe ∆H for the reaction, as written, is -6542 kJ. This represents the heat generated from burning 2 moles of C6H6(l) and generating 12 moles of CO2(g), etc.From the mass of C6H6 given, we can calculate moles of C6H6 combusted. From that value, we can then find the heat that would be generated. After finding the heat generated, we can find the change in temperature of the water.5.300 g C6H6 x 1 mol/78.11 g = 0.06785 molesheat generated = 0.06785 moles x 6542 kJ/2 moles = 221.9 kJq = mC∆T221.9 kJ = (5.691 kg)(4.184 kJ/kg/deg)(∆T)

2C6H6(l) + 15O2(g) --> 12CO2(g) + 6H2O(l) ... balanced equationThe ∆H for the reaction, as written, is -6542 kJ. This represents the heat generated from burning 2 moles of C6H6(l) and generating 12 moles of CO2(g), etc.From the mass of C6H6 given, we can calculate moles of C6H6 combusted. From that value, we can then find the heat that would be generated. After finding the heat generated, we can find the change in temperature of the water.5.300 g C6H6 x 1 mol/78.11 g = 0.06785 molesheat generated = 0.06785 moles x 6542 kJ/2 moles = 221.9 kJq = mC∆T221.9 kJ = (5.691 kg)(4.184 kJ/kg/deg)(∆T)∆T = 221.9 kJ/(5.691 kg)(4.184 kJ/kg/deg)

2C6H6(l) + 15O2(g) --> 12CO2(g) + 6H2O(l) ... balanced equationThe ∆H for the reaction, as written, is -6542 kJ. This represents the heat generated from burning 2 moles of C6H6(l) and generating 12 moles of CO2(g), etc.From the mass of C6H6 given, we can calculate moles of C6H6 combusted. From that value, we can then find the heat that would be generated. After finding the heat generated, we can find the change in temperature of the water.5.300 g C6H6 x 1 mol/78.11 g = 0.06785 molesheat generated = 0.06785 moles x 6542 kJ/2 moles = 221.9 kJq = mC∆T221.9 kJ = (5.691 kg)(4.184 kJ/kg/deg)(∆T)∆T = 221.9 kJ/(5.691 kg)(4.184 kJ/kg/deg)∆T = 9.3 degrees

2C6H6(l) + 15O2(g) --> 12CO2(g) + 6H2O(l) ... balanced equationThe ∆H for the reaction, as written, is -6542 kJ. This represents the heat generated from burning 2 moles of C6H6(l) and generating 12 moles of CO2(g), etc.From the mass of C6H6 given, we can calculate moles of C6H6 combusted. From that value, we can then find the heat that would be generated. After finding the heat generated, we can find the change in temperature of the water.5.300 g C6H6 x 1 mol/78.11 g = 0.06785 molesheat generated = 0.06785 moles x 6542 kJ/2 moles = 221.9 kJq = mC∆T221.9 kJ = (5.691 kg)(4.184 kJ/kg/deg)(∆T)∆T = 221.9 kJ/(5.691 kg)(4.184 kJ/kg/deg)∆T = 9.3 degreesFinal temperature = 21º + 9.3º = 30.3ºC

Answer 2

The values of 'm' and 'n' in the reaction $\[2{C_6}{H_6} + 15{O_2} \to mC{O_2} + n{H_2}O\]$ are 12 and 6 respectively.

Explanation:

• A balanced reaction is a reaction in which the number of atoms of all elements of the reactants on the reactant side is equal to the number of atoms of all elements of the products on the product side.
• To balance the given chemical equation, all the atoms of all the elements on both sides must be made equal.
• To balance C atoms, put 12 as the coefficient before $\[C{O_2}\]$. The equation becomes $\[2{C_6}{H_6} + 15{O_2} \to 12C{O_2} + n{H_2}O\]$
• Now, put 6 as the coeffiecient before $\[{H_2}O\]$ to balance O and H atoms.
• Therefore, the balanced chemical equation obtained is $\[2{C_6}{H_6} + 15{O_2} \to 12C{O_2} + 6{H_2}O\]$
• Thus, according to the balanced equation, the values of 'm' and 'n' are 12 and 6 respectively.