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

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

Explanation:

For equation (I): ⇒ (0.22 moles of )

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)

= atoms

For equation (II): ⇒

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 = atoms.

For equation (III):  ⇒

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 = atoms.

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

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