This is a limiting reagent problem. How do I know that. Because BOTH reactants are given. Basically, what we do is work two simple stoichiometry problems, identify the limiting reagent as the one producing the smaller amount of product.
CH4(g) + 2 O2(g) => CO2(g) + 2 H2O(g).
First CH4: Using the coefficients in the balanced equation, convert 17 moles CH4 to moles H2O.
17 moles CH4 x (2 moles H2O/1 mole CH4) = 34 mols H2O
Next O2: Same process.
7 mols O2 x (2 moles H2O/2 mols O2) = 7 moles H2O.
Obviously both answers can be right which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value; therefore, the reaction will produce 7 moles H2O and O2 is the limiting reagent.
Now convert 7 moles H2O to grams.
g = mols x molar mass.
Consider the reaction,
CH4(g) + 2 O2(g) => CO2(g) + 2 H2O(g).
How many grams of H2O(g) will be made by the reaction of 17 moles of CH4(g) with 7 moles of O2(g)?
2 answers
thanks..now I know how to do this kind of problem...you rock!