When 16 grams of methane gas combine with 64 grams of oxygen and 44 grams of carbon dioxide plus water what mass of water is produced?

Answer 1

1 mol Hydrogen (H) masses 1 gram

1 mol Oxygen (O) masses 16 gram

2 H2 + O2 --> 2H2O

Equiv is H2 + 1/2 O2 --> H2O

that is 1 mol of water

2g +16g ----> 18g H2O

This reaction is explosive and goes to completion under standard temperature and pressure, but for pedants, all reactions are dynamic equilibria so in a confined volume there will be the tiniest proportion of H2 and O2 remaining. So minutely less than 18g is the answer.

Answer 2

First you need to write a balanced chemical equation. You are given that methane is burned, meaning a combustion reaction in which carbon dioxide and water are released.

Unbalanced: CH4 + O2 ---> CO2 + H2O

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

Givens:

16.0 grams CH4 (Molecular mass 16.0 grams)

64.0 grams O2 (Molecular mass 32.0 grams)

44.0 grams H2O (Molecular mass 18.0 grams)

Mole ratio: 1:2:1:2 (CH4:O2:CO2:H2O)

Then you need to find which of the reactants are the limiting reactant and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.

16.0 g CH4 / (16.0 g) = 1.00 moles CH4

64.0 g O2 × (2 moles) / (32.0 g) = 4.00 moles O2

Because methane is the limiting reactant, the amount of water produced can only be as much as the methane. Take the amount of methane and use stoichiometry to find the amount of water produced in grams.

16.0 g CH4 / (16.0 g CH4) × (2 moles O2) × (18.0 g O2) = 36.0 grams H2O

Answer 3

Since water and carbon dioxide are the only products of this reaction, the answer can be found from the law of conservation of mass. The water mass must equal the sum of the masses of the reactants minus the mass of the other product formed.

24 + 96 - 66 = 54 grams of water.

Answer 4

The balanced equation for the reaction is O2 + 2 H2 -> 2 H2O. The gram molecular mass of O2 is 2 X 15.9994, and the gram molecular mass of water is 18.015928. Therefore, the mass of water produced will have a ratio to the mass of oxygen reacted of [2(18.015928)/2(15.9994)] or about 1.126, and the amount of water produced will be 18.5 grams, to the justified number of significant digits.

Answer 5

Step 1) Determine limiting reagent.

Reaction: 2H + 1O-->1H20.

24 g O*1 mol O/MW of O*(1 H20/1 O)= moles of H20 possible*MW of H20=mass of H20 possible

16 g H*1 mole of H/MW of H*(1 H20/2 H)= moles of H20 possible*MW of H20=mass of H20 possible

Whichever amount yields the smalled amount is the limiting reagent. Therefore, the corresponding moles of H20 is the max mass of water you can make.

Answer 6

CH4 + 2O2 -> CO2 + 2H2O

moles of H2O produced = 36/18 = 2

moles of CO2 produced = 44/44 = 1

moles of CH4 reacted = 16/16 = 1

so moles of O2 consumed = 2 i.e 2 X 32 g = 64 g of O2

Answer 7

Assuming that the last word of the question should have been "needed" rather than "nee" and that carbon dioxide and water are the only products of the reaction, the answer can be found from the Law of Conservation of Mass during chemical reactions:

36 + 44 - 16 = 64 grams of oxygen needed.

Answer 8

2 grams of hydrogen gives 1 mole of hydrogen gas

16 grams of oxygen gives 1/2 mole of oxygen gas.

1H2+1/2 O2--->1H2O

1 mole of water=1x(2+16)

=18g

Answer 9

36 grams

other websites help :P

Answer 10

We need 64 g oxygen.