Consider the combustion of methane: CH_(4)+2O_(2)arrow CO_(2)+2H_(2)O If 16 grams of methane (molar mass of 16grams/mol) reacts with 32 grams of oxygen (molar mass of 32 grams per mol), what is the limiting reagent and how much methane is combusted? methane is the limiting reagent and it is all reacted oxygen is the limiting reagent and all of the me/yane reacts oxygen is the limiting reagent and a little more than 4 grams of methane reacts oxygen is the limiting reagent and 8 grams of methane reacts

To determine the limiting reagent and the amount of methane combusted, we need to compare the mole ratio of methane to oxygen in the reaction.<br /><br />Given information:<br />- Molar mass of methane (CH4) = 16 g/mol<br />- Molar mass of oxygen (O2) = 32 g/mol<br />- Mass of methane = 16 g<br />- Mass of oxygen = 32 g<br /><br />Step 1: Calculate the number of moles of methane and oxygen.<br />Number of moles of methane = Mass of methane / Molar mass of methane<br />Number of moles of methane = 16 g / 16 g/mol = 1 mol<br /><br />Number of moles of oxygen = Mass of oxygen / Molar mass of oxygen<br />Number of moles of oxygen = 32 g / 32 g/mol = 1 mol<br /><br />Step 2: Determine the limiting reagent.<br />The balanced chemical equation for the combustion of methane is:<br />CH4 + 2O2 → CO2 + 2H2O<br /><br />From the balanced equation, we can see that 1 mol of methane reacts with 2 mol of oxygen.<br />Since we have 1 mol of methane and 1 mol of oxygen, oxygen is the limiting reagent.<br /><br />Step 3: Calculate the amount of methane combusted.<br />Since oxygen is the limiting reagent, all of the methane will react.<br />Therefore, the amount of methane combusted is 1 mol, which is equivalent to 16 g.<br /><br />Therefore, the correct answer is:<br />oxygen is the limiting reagent and all of the methane reacts.