Electrones de Rh 103 , Be9, CE 140 Au197, K39, CF 251, V 51?

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Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

For  :-

Mass of  = 1.28 g

Molar mass of  = 16.04 g/mol

The formula for the calculation of moles is shown below:

Thus,

For  :-

Mass of  = 10.1 g

Molar mass of  = 31.998 g/mol

The formula for the calculation of moles is shown below:

Thus,

According to the given reaction:

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

Limiting reagent is the one which is present in small amount. Thus, is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

Thus,

Mass of = 3.51 g

Theoretical yield = 3.51 g