Lets write the reaction equation first: CH 4 + 2O 2 ----->CO 2 + 2H 2 O moles of

Question

Lets write the reaction equation first:

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

moles of methane is = mass /molar mass = 3.69g / 16g/mol = 0.231 mol

moles of oxygen is = 22g / 32/mol = 0.6875 mol

1 mol CH4 needs 2 mol O2 to react completly

0.231 mol CH4 needs 2 x 0.231 = 0.462 moles O2

but we have 0.6875 moles O2 , which is in excess and CH4 becomes limiting reactant here

The product formed is given by limiting reactant only

1 mol CH4 froms 2 mol H2O

0.231 mol CH4 will form 2 x 0.231 mol = 0.462 mol H2O

mass of H2O = molar mass x moles = 0.462 mol x 18.01 g/mol = 8.32 g

Ans = 8.32 g

Explanation / Answer

O STOICHIOME TRY Limiting reactants Gaseous methane (CH.) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO2) and gaseous water (H,0) Suppose 3.69 g of methane is mixed with 22. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits Gaseous methane (CH4) will react with

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