Question

iven the following chemical equations:I. C+ 2H_2 2​ → CH_4 4​ (0.22 moles of H_2 2​ )II. 6CO_2 2​ + 6H_2 2​ O → C_6 6​ H_{12} 12​ O_6 6​ (2.25 g of CO_2 2​ )III. CH_4 4​ + 2O_2 2​ → CO_2 2​ + 2H_2 2​ O (1.25 L of CH_4 4​ with density = 0.656 g/L)Which equation will have the highest number of carbon atoms produced based on the information given next to each chemical equation

Answer 1

Answer:

Equation (I) produced the highest number of carbon atoms.

Explanation:

For equation (I): $C + 2H_{2}$⇒$CH_{4}$ (0.22 moles of $H_{2}$)

2 moles of hydrogen gas will produce 1 mole of methane gas. Therefore, 0.22 moles of hydrogen gas will produced (0.22/2) = 0.11 moles of methane gas.

Similarly, 1 mole of methane gas has one mole of carbon atom and 4 moles of hydrogen atoms. Therefore, 0.11 moles of methane gas will have 0.11 mole of carbon atom.

The number of carbon atom = (moles of carbon atom)*(Avogadro's number)

= $0.11*(6.02*10^{23}) = 6.622*10^{22}$ atoms

For equation (II): $6CO_{2}+6H_{2}O$⇒$C_{6}H_{12}O_{6}$

2.25 g of carbon dioxide gas = [2.25g/(44g/mol)] = 0.051 mole of carbon dioxide gas.

In equation (II), 6 moles of carbon dioxide gas produced 1 mole of glucose. Therefore, 0.051 moles of carbon dioxide gas will produce (0.051/6) = 0.0085 moles of glucose.

Similarly, 1 mole of glucose contains 6 moles of carbon atom. Therefore, 0.0085 moles of glucose will contain (0.0085*6) = 0.051 moles of carbon atom.

The number of carbon atom = $0.051*6*10^{23} = 3.07*10^{22}$ atoms.

For equation (III):  $CH_{4}+2O_{2}$⇒$CO_{2}+2H_{2}O$

1.25 L of methane gas with density of 0.656 g/L has a mass of (1.25 L*0.656 g/L) = 0.82 g of methane gas.

moles of methane gas = 0.82g/(16g/mol) = 0.051 mol

In equation (III), 1 mole of methane gas produced 1 mole of carbon dioxide gas. Therefore, 0.051 moles of methane gas produced 0.051 moles of carbon dioxide gas.

Similarly, 1 mole of carbon dioxide gas has one mole of carbon atom. Thus, 0.051 moles of carbon dioxide gas will have 0.051 moles of carbon atom.

The number of carbon atoms = $0.051*6*10^{23} = 3.08*10^{22}$ atoms.

Therefore, Equation (I) produced the highest number of carbon atoms.