If 16g of methane is allowed to react w 32.0g of O2 what is the limiting reagent?

Question

fire, water, carbon dioxide, methane, or oxygen?????

Answer 1

16 g CH4 = 1 mol CH4 since C =12 and H = 1 x 4

32 g O2 = 1 mol O2 since O = 16 x 2

write and balance the equation

CH4 + 2 O2 ---> CO2 + 2 H2O

The ratio of CH4 to O2 is 1 : 2

you need twice as many moles of O2 as CH4, so O2 is the limiting reactant.

Answer 2

So first off you need the balanced equation for this reaction which is

CH4 +2O2 --> CO2 + 2H2O

So, the equation tells us that for every one molecule of methane that reacts, 2 molecules of oxygen gas are required. Similarly, 2 moles of oxygen gas are required to react with just one mole of methane.

Now, CH4 has a molar mass of 16 g/mol whereas O2 has a molar mass of 32 g/mol. You have one mole of both reactant implying that oxygen will be your limiting reagent since you need 2 moles, not 1 mole, of oxygen gas to react with all 16 g of methane.

Answer 3

First balance the combustion equation, it only goes one way because it is combustion, and cannot be at equilibrium:

1 CH4 + 2 O2 -> 1 CO2 + 2 H2O + energy

then find the molar masses of the reactants using the periodic Table:

CH4 = 16.05 g/mole
O2 = 32.00 g/mole

Use dimensional analysis to find out how many moles reacted in
the equation:

CH4 16g divide by 16.05g/mole = 1 mole
O2 32g divide by 32.00 g/mole = 1 mole

32g of O2 was used in the reaction, or 1 mole right? If we look at the balaced equation, it requires "2" moles of O2 to form the products. for every 1 mole (or16g) of CH4.This means that you will run out of O2 before you run out of CH4, making it the limiting reagent.

*Always balance the equations first, then look at how many moles of products were used in the reaction, and whichever mole quantity is the least, will give you your limiting reagent.

Answer 4

The reaction is 4NH3(g) + 5O2(g) -------> 4NO(g) + 6H2O(l) mass NH3 presented = 21.4 g mole NH3 = 21.4 g/(1mol /17.03 g) = a million.257 mol mass O2 presented = 40 two.5 g mole O2 = 40 two.5 g(1mol/32 g) = a million.328 mol mole O2 mandatory for a million.257 mol NH3 = a million.257 mol NH3 (5 mol O2/4 mol NH3) = a million.571 mol a million.571 mol O2 is mandatory yet in basic terms a million.328 mole is provided, as a result NH3 is extra and O2 is the proscribing reactant.