Question

Find the moles of h2o produce when 5.8g of butane is burn also find the molecules of o2 required and the atoms present in the h2o that produce at given condition also find the total no. Of atoms produce in the product 2c4h10 + 13o2 + 10h2o + 8co2

Answer 1

Moles of H2O = 0.5 mol, Molecules of O2 required = 3.9143 x 10^23, Atoms in H2O = 9.033 x 10^23, Total atoms in products = 1.62594 x 10^24

Answer 2

The problem involves stoichiometry based on the combustion of butane. We are given the balanced chemical equation and the mass of butane. We need to find several quantities related to the reactants and products.

The balanced chemical equation for the combustion of butane is: $2C_4H_{10} + 13O_2 \rightarrow 10H_2O + 8CO_2$

1. Calculate moles of Butane ($C_4H_{10}$): Molar mass of C = 12 g/mol Molar mass of H = 1 g/mol Molar mass of $C_4H_{10} = (4 \times 12) + (10 \times 1) = 48 + 10 = 58$ g/mol. Given mass of butane = 5.8 g. Moles of butane = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{5.8 \text{ g}}{58 \text{ g/mol}} = 0.1 \text{ mol}$.

2. Moles of H₂O produced: From the balanced equation, 2 moles of $C_4H_{10}$ produce 10 moles of $H_2O$. Using mole ratio: Moles of $H_2O = 0.1 \text{ mol } C_4H_{10} \times \frac{10 \text{ mol } H_2O}{2 \text{ mol } C_4H_{10}} = 0.1 \times 5 = 0.5 \text{ mol } H_2O$.

3. Molecules of O₂ required: From the balanced equation, 2 moles of $C_4H_{10}$ require 13 moles of $O_2$. Using mole ratio: Moles of $O_2 = 0.1 \text{ mol } C_4H_{10} \times \frac{13 \text{ mol } O_2}{2 \text{ mol } C_4H_{10}} = 0.1 \times 6.5 = 0.65 \text{ mol } O_2$. To find the number of molecules, multiply by Avogadro's number ($N_A = 6.022 \times 10^{23}$ molecules/mol): Molecules of $O_2 = 0.65 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 3.9143 \times 10^{23}$ molecules.

4. Atoms present in the H₂O produced: Moles of $H_2O$ produced = 0.5 mol. Number of $H_2O$ molecules = $0.5 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 3.011 \times 10^{23}$ molecules. Each $H_2O$ molecule contains 2 hydrogen atoms and 1 oxygen atom, totaling 3 atoms per molecule. Total atoms in $H_2O = 3.011 \times 10^{23} \text{ molecules} \times 3 \text{ atoms/molecule} = 9.033 \times 10^{23}$ atoms.

5. Total number of atoms produced in the products (H₂O and CO₂): First, calculate moles of $CO_2$ produced. From the balanced equation, 2 moles of $C_4H_{10}$ produce 8 moles of $CO_2$. Using mole ratio: Moles of $CO_2 = 0.1 \text{ mol } C_4H_{10} \times \frac{8 \text{ mol } CO_2}{2 \text{ mol } C_4H_{10}} = 0.1 \times 4 = 0.4 \text{ mol } CO_2$. Number of $CO_2$ molecules = $0.4 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 2.4088 \times 10^{23}$ molecules. Each $CO_2$ molecule contains 1 carbon atom and 2 oxygen atoms, totaling 3 atoms per molecule. Total atoms in $CO_2 = 2.4088 \times 10^{23} \text{ molecules} \times 3 \text{ atoms/molecule} = 7.2264 \times 10^{23}$ atoms.

Total atoms in products = Total atoms in $H_2O$ + Total atoms in $CO_2$ Total atoms in products = $9.033 \times 10^{23} \text{ atoms} + 7.2264 \times 10^{23} \text{ atoms}$ Total atoms in products = $(9.033 + 7.2264) \times 10^{23} \text{ atoms} = 16.2594 \times 10^{23} \text{ atoms}$ Total atoms in products = $1.62594 \times 10^{24}$ atoms.

Summary of Results:

• Moles of H₂O produced: 0.5 mol
• Molecules of O₂ required: $3.9143 \times 10^{23}$ molecules
• Atoms present in the H₂O produced: $9.033 \times 10^{23}$ atoms
• Total number of atoms produced in the products (H₂O and CO₂): $1.62594 \times 10^{24}$ atoms