How many liters of oxygen are needed to exactly react with 23.8 g of methane at STP?

How many liters of oxygen are needed to exactly react with 23.8 g of methane at STP?

66.483L
Therefore, 66.483L of oxygen are needed to exactly react with 23.8 g of methane at STP.

How many liters of oxygen are needed to exactly react with 21.8 g of methane at STP?

FREE Expert Solution. We are asked to calculate the volume of oxygen gas (in liters) needed to react with 21.8 g of methane at STP. The volume of oxygen gas that reacted is 66.7 L.

How many grams of oxygen is required for complete combustion of methane?

The molar mass of oxygen gas is 32 grams/mole while the molar mass of methane is equal to 16.042 grams/mole. The mass of oxygen gas needed for the complete combustion of 25.5 grams of methane is 102 grams.

How many liters of O2 are needed to react completely?

Since H2 reacts with O2 in a 2:1 ratio, to completely react the two you will need half as much oxygen as hydrogen. To find liters of oxygen, just multiply the moles above by 22.4 liters.

What volume of oxygen are needed to exactly react with 25.8 g of methane?

Explanation: And so we gots the combustion equation… And so moles of methane=25.8⋅g16.04⋅g⋅mol−1=1.536⋅mol ..

How many grams of oxygen is needed for complete combustion of 6g CH4?

So 16 g of Methane requires 64 g of oxygen for the complete combustion.

What volume of hydrogen will react with 22.4 liters of oxygen to form water all volumes are measured at STP?

2 × 22.4 liters of hydrogen react with 22.4 liters of oxygen. The volume of hydrogen that will react is 44.8 liters.

How many liters of o2 are consumed during the combustion of 60.0 grams of ethane at STP in the open atmosphere?

1 mole of any gas occupies 22.4litres. For 60g of ethane we need= (78.4/30)x 60=156.8 L of oxygen.

Does methane react with oxygen?

When methane (CH4) reacts with oxygen it forms carbon dioxide and water. Here we notice that the reaction is not balanced as the number of hydrogen atoms are different on each side. The same is true for the number of oxygen atoms.

How many moles of oxygen are needed to react with CH4?

However, in the reaction equation, we need 2 moles of oxygen for every one mole of Methane, and we only have 1 mole of oxygen and 2 moles of Methane. That means methane is in excess and oxygen is thus our limiting reactant. So in this reaction we would only be able to react 1 mol of O2 with 0.5 mol of CH 4.

What is the limiting reactant in the combustion of methane?

Consider the combustion of methane: Ch4 (g)+2O2 (g) –> CO2 (g)+2H2O (g), suppose 2.8 moles of methane are allowed to react with 3 moles of oxygen, what is the limiting reactant? O2 is the limiting reactant. There are different methods of finding the limiting reactant, here is the simplest to my opinion.

What is your limiting reactant if 32g of CH4 reacts with oxygen?

What is your limiting reactant if 32 grams of CH4 reacts with 32 grams of oxygen gas? Oxygen is our limiting reactant. We must start by creating a balanced reaction between Methane and Oxygen. As methane is a hydrocarbon reacting with oxygen this will be a combustion reaction resulting in carbon dioxide and water.

How do you find the limiting reactant of a reaction?

There are different methods of finding the limiting reactant, here is the simplest to my opinion. Find the molar ratio between the experimental number of moles and theoretical number of moles of the reactants: For CH 4: 2.8mol 1mol = 2.8. The 1 mol is taking from the coefficient of CH 4 from the balanced equation.