Question

What mass of butane in grams is necessary to produce 1.5×103 kj of heat what mass of co2 is produced?

Answer 1

$\boxed{32.7{\text{3 g}}}$ of butane is necessary to produce $1.5 \times {10^3}\;{\text{kJ}}$of heat and $\boxed{99.32\;{\text{g}}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ is produced.

Further explanation:

Stoichiometry:

It is used to determine the amount of species present in the reaction by the relationship between reactants and products. It is used to determine the moles of a chemical species when moles of other chemical species present in the reaction is given.

Consider the general reaction,

${\text{A}} + 2{\text{B}} \to 3{\text{C}}$

Here,

A and B are reactants.

C is the product.

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Combustion reaction:

It is the reaction in which the reactant reacts with molecular oxygen to form carbon dioxide and a water molecule. Molecular oxygen acts as the oxidizing agent in these reactions. A large amount of heat is released and therefore combustion reactions are exothermic in nature.

The combustion of butane occurs as follows:

${{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}+\frac{{13}}{2}{{\text{O}}_2} \to 4{\text{C}}{{\text{O}}_2} + 5{{\text{H}}_{\text{2}}}{\text{O}}$

The value of $\Delta {{\text{H}}_{{\text{reaction}}}}$ is -2658 kJ/mol. This indicates the heat produced when one mole of butane is combusted.

The number of moles of butane $\left( {{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}} \right)$ required in the given reaction is calculated as follows:

$\begin{aligned}{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}&=\left({-150{\text{0 kJ}}} \right)\left( {\frac{{{\text{1 mol}}}}{{ - 265{\text{8 kJ}}}}}\right)\\&=0.56{\text{43 mol}} \\ \end{aligned}$

The formula to calculate the mass of butane $\left({{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}\right)$ is as follows:

${\text{Mass of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}=\left( {{\text{Moles of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}\right)\left({{\text{Molar mass of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}}\right)$                   ...... (1)

The number of moles of ${{\text{C}}_4}{{\text{H}}_{10}}$ is 0.5643 mol.

The molar mass of ${{\text{C}}_4}{{\text{H}}_{10}}$ is 58 g/mol.

Substitute these values in equation (1).

$\begin{aligned}{\text{Mass of }}{{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}&=\left( {{\text{0}}{\text{.5643 mol}}}\right)\left( {\frac{{{\text{58 g}}}}{{{\text{1 mol}}}}}\right)\\&=32.72{\text{94 g}}\\&\approx{\text{32}}{\text{.73 g}} \\ \end{aligned}$

Therefore the mass of butane required is 32.73 g.

According to the reaction stoichiometry, one mole of ${{\text{C}}_4}{{\text{H}}_{10}}$ produces four moles of ${\text{C}}{{\text{O}}_{\text{2}}}$. The moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ produced in the given reaction are calculated as follows:

$\begin{aligned}{\text{Moles of C}}{{\text{O}}_{\text{2}}}&=\left({{\text{0}}{\text{.5643 mol }}{{\text{C}}_4}{{\text{H}}_{10}}}\right)\left({\frac{{{\text{4 mol C}}{{\text{O}}_{\text{2}}}}}{{1\;{\text{mol }}{{\text{C}}_4}{{\text{H}}_{10}}}}} \right)\\&=2.257{\text{2 mol}}\\\end{gathered}$

The formula to calculate the mass of ${\text{C}}{{\text{O}}_{\text{2}}}$ is as follows:

${\text{Mass of C}}{{\text{O}}_{\text{2}}} = \left( {{\text{Moles of C}}{{\text{O}}_{\text{2}}}} \right)\left( {{\text{Molar mass of C}}{{\text{O}}_{\text{2}}}} \right)$              ...... (2)

Substitute 2.2572 mol for the moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ and 44 g/mol for the molar mass of ${\text{C}}{{\text{O}}_{\text{2}}}$ in equation (2).

$\begin{aligned}{\text{Mass of C}}{{\text{O}}_{\text{2}}}&=\left({{\text{2}}{\text{.2572 mol}}} \right)\left( {\frac{{{\text{44 g}}}}{{{\text{1 mol}}}}} \right)\\&=99.316{\text{8 g}}\\&\approx {\text{99}}{\text{.32 g}} \\ \end{aligned}$

Therefore the mass of carbon dioxide produced is 99.32 g.

Learn more:

1. Calculate $\Delta {\text{H}}$ for the reaction using Hess law: brainly.com/question/11293201

2. Calculate the hydroxide ion concentration: brainly.com/question/11293214

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: butane, mass, CO2, C4H10, molar mass of CO2, molar mass of C4H10, 32.73 g, 99.32 g, 44 g/mol, 58 g/mol, moles of CO2, moles of C4H10, 2.2572 mol, 0.5643 mol, stoichiometry, one mole, four moles.

Answer 2

The heat of reaction (i.e. combustion) of butane ($C_{4} H_{10}$) when reacted with oxygen ($O_{2}$)  is -2658 kJ/mol butane, and the chemical reaction is given by: 

$C_{4} H_{10}$ + $\frac{13}{2} O_{2}$ ---> $4 CO_{2}$  + $5 H_{2}O$

The mass of butane required in the reaction is based on the heat produced by the reaction, which is given to be -1,500 kJ. The minus sign is added because the reaction releases heat (exothermic), which means that the products are in a "lower energy state" than the reactants. 

Dividing this with the heat of reaction per mole of butane reacted would give the number of moles butane required. Then, multiplying the answer with the molar mass of butane which is 58 grams/mole, will give the mass of butane required. 

Moles of butane = [(-1,500 kJ)/(-2658 kJ/mol butane)]
Moles of butane = 0.5643 moles butane

Mass of butane  = 0.5643 moles butane * 58 grams/mol butane
Mass of butane  = 32.73 grams butane

The mass of carbon dioxide ($CO_{2}$) can be determined by multiplying the moles of butane ($C_{4} H_{10}$) with the mole ratio of ($CO_{2}$) produced to the ($C_{4} H_{10}$) reacted, and then with the molar mass of ($CO_{2}$), which is 44 grams/mole. 

Mass of carbon dioxide produced 
    = 0.5643 moles butane * [4 moles $CO_{2}$/ 1 mole $C_{4} H_{10}$] * 44 grams/mole $CO_{2}$

Mass of carbon dioxide produced  
    = 99.32 grams $CO_{2}$

Thus, the mass of butane required is 32.73 grams, and the mass of carbon dioxide produced from the reaction of this amount of butane is 99.32 grams.