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Use the equation below to solve the problem that follows.

2H2 (g) + O2 (g) → 2H2O (g)

When David reacts 13.8 grams of hydrogen gas with excess oxygen, 87.0 grams of water are formed. Calculate his percent yield of water.

Im very loss with this question

Respuesta :

ardni313

Percent yield of water : 70%

Further explanation

Reaction

2H₂ (g) + O₂ (g) → 2H₂O (g)

mol H₂(MW= 2 g/mol) :

[tex]\tt \dfrac{13.8~g}{2~g/mol}=6.9~mol[/tex]

mol H₂O = mol H₂ = 6.9

mass H₂O(theoretical) :

[tex]\tt 6.9\times 18~g/mol=124.2~g[/tex]

[tex]\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{87}{124.2}\times 100\%=70\%[/tex]

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