Question

Use 1 decimal point for all atomic masses. 18 g of CH_4(g) are reacted with 5.2 g of O_2(g) by the following reaction

CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)

What is the limiting reagent?

*My Answer: CH4O2

Based on the limiting reagent, what should the yield of H2O(g) be? in ________GRAMS

I am getting the grams wrong...can you help?

Answer 1

The balanced chemical equation is ---

CH4(g) + 2O2(g) = CO2(g) + 2H2O(g)

From the reaction equation we can see 1 mole CH4 reacted with 2 moles of O2 to produce 2 moles of H2O.

Molar mass of CH4 is 16 g/mole

Mass of CH4 present = 18 g

So number of moles of CH4 present = (18/16) moles = 1.125 moles

Molar mass of O2 = 32 g/mole

Mass of O2 present= 5.2 g

So number of moles of O2 present = (5.2/32) moles = 0.1625 moles

Now limiting reagent in a chemical reaction is the reagent which used up first and determines the amount of product will be produced.

Here we can see O2 will be used up first, thus O2 is the limiting reagent here.

Now from 2 moles of O2 number of moles of H2O produced is 2 moles.

So from 0.1625 moles of O2 number of moles of H2O will be produced= 0.1625 moles

Now molar mass of H2O is 18 g/mole

So mass of H2O present in 0.1625 moles of H2O is = (0.1625×18) g = 2.925 g

So 2.925 grams of H2O will be produced.(answer)