Determine (A) the limiting reactant; and (B) the mass (in g) of H_2O if 50.0g of

Question

Determine (A) the limiting reactant; and (B) the mass (in g) of H_2O if 50.0g of methane (CH_4) and 75.0 g of oxygen react according to the following balanced reaction. And finally, determine (C) the percent yield if 40.0 g of H_2O are produced in the reaction. (Possibly useful molar masses: CH_4 = 16.05 g/mol; H_2O = 18.02 g/mol; O_2 = 32.00 g/mol). CH_4 (g) + 2 O_2 (g) rightarrow CO_2 (g) + 2 H_2O (g) A. Circle the limiting reactant CH_4 O_2 Determine the Theoretical yield (mass) of H_2O: Answer: _____ Determine the percent yield of H_2O: Answer: _____

Explanation / Answer

A. limiting reactant; O2

as per reaction CH4 moles = weight /molar mass

= 50/16.05

=3.115 moles

O2 moles = weight /molar mass

= 75/32.0

=2.34 moles

as based on stachiometry reaction , O2 moles require double (2O2) with respect to CH4 , here 3.11 moles of CH4 but O2 is 2.34 moles only (require 3.115*2 =6.23 moles). so the limiting reactant is O2

B. theoretical yield (mass) of H2O :

as based on above calculation the limiting reactant O2 (75 g)

this 75 g of O2 produce how much water

= weight/molarmass*water molar mass

=75/32.0*18

=42.18 g of water.

so theoretically 75 g of O2 should produce 42.18 g of water.

C. Percent of yield of H2O:

  principle= isolated yield/theoretical yield*100

= isolated weight/ theoretical weight*100

as mentioned reaction 40 g produced, but theoretical (above calculation) weight 42.18 g

so = 40/42.18*100

=94.83%

the percent of H2O yield 94.83%

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