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If you produce 10 g of water in the reaction from 20 g of O2 what is your percent yield? 2H2+O2>2H2O

Sagot :

jbferrado

Answer:

44.4%

Explanation:

The balanced chemical equation for the reaction is:

2 H₂(g) + O₂(g) → 2 H₂O(g)

According to the equation, 1 mol of O₂ produces 2 moles of water (H₂O). We convert the moles to grams by using the molecular weight (Mw) of each compound:

Mw(O₂) = 2 x 16 g/mol O = 32 g/mol

mass O₂ = 1 mol x 32 g/mol = 32 g O₂

Mw(H₂O) = (2 x 1 g/mol H) + 16 g/mol O = 18 g/mol

mass H₂O = 2 mol x 18 g/mol = 36 g H₂O

Thus, from 32 grams of O₂, 36 g of H₂O are produced (stoichiometric ratio= 36 g H₂O/32 g O₂). We calculate the amount of H₂O that would be produced from 20 grams of O₂ by multiplying the stoichiometric ratio by the mass of O₂, as follows:

36 g H₂O/32 g O₂ x 20 g O₂ = 22.5 g H₂O

The theoretical yield of H₂O is 22.5 g, and the actual amount produced is 10 g. So, we calculate the percent yield as follows:

Percent yield = actual yield/ theoretical yield x 100

                      = (10 g)/(22.5 g) x 100

                      = 44.4%

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