We are provided an equation,CH4+ 2O2CO2+ 2H2O, and asked come answer several questions:

Part A.

You are watching: Ch4 + 2o2 co2 + 2h2o

We room asked to determine thelimiting reagent, given the variety of moles of each reactant (2.55 moles of methane and 4.50 moles of oxygen).

Part B.We room asked come calculate how manygrams the products(carbon dioxide and water) are created from the mixture in component A.

Part C.We are asked to calculate theactual yield, offered the percent yield, 95.0%.

Part D.We space asked come calculate exactly how manygrams that the excess(if over there is any) will continue to be unreacted.

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Problem Details

Given this equation: CH4 + 2O2 → CO2 + 2H2O

a) If 2.55 moles of methane reacts v 4.50 mole of oxygen, what is the limiting reagent?

b) indigenous this mixture that reactants, how numerous grams the carbon dioxide and also how plenty of grams that water can be produced?

c) If the percent yield because that this reaction is 95.0%, what really yield deserve to you expect?

d) How numerous grams of the excess reagent (if there is one) will remain unreacted?

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