H2 + O2 =H2O A. calculate the maximum number of grams of water that can beproduc

Question

H2 + O2 =H2O A. calculate the maximum number of grams of water that can beproduced from 5.65g of hydrogen gas and 3.44 g of oxygen gas. B. Which is the limiting reactant? C. which is the excess reactant? D. If 2.50 g of H2O is the actual yeild,what is the percentyeild of the reaction? H2 + O2 =H2O A. calculate the maximum number of grams of water that can beproduced from 5.65g of hydrogen gas and 3.44 g of oxygen gas. B. Which is the limiting reactant? C. which is the excess reactant? D. If 2.50 g of H2O is the actual yeild,what is the percentyeild of the reaction?

Explanation / Answer

2H2 + O2 -------->2H2O Molar mass of H2O is = 2 * 1 + 16 = 18 g 2 * 2 g of Hydrogen on reaction with 2 * 16 g of Oxygenproduces 2 * 18 g of water (i.e., )4g of Hydrogen on reaction with32 g of Oxygenproduces 36 g of water A. calculate the maximum number of grams of water that can beproduced from 5.65g of hydrogen gas and 3.44 g of oxygen gas. 32 g of Oxygen reacts with 4 g of Hydrogen 3.44 g of Oxygen reacts withX g of Hydrogen X = ( 4 * 3.44 ) / 32     = 0.43 g So, 5.65 - 0.43 = 5.22 g of H2 remain unreacted 32 g of Oxygen produces 36 g of water 3.44 g of  Oxygen produces Y g of water Y = ( 36 * 3.44 ) / 32     = 3.87 g of water B. Which is the limiting reactant? Since the H2 mass is than the required amount , it is thelimiting reactant. C. which is the excess reactant? D. If 2.50 g of H2O is the actual yeild,what is the percentyeild of the reaction? % yield = (actual yield / Theoritical yield ) * 100            = ( 2.5 / 3.87 ) * 100            = 64.6 % A. calculate the maximum number of grams of water that can beproduced from 5.65g of hydrogen gas and 3.44 g of oxygen gas. 32 g of Oxygen reacts with 4 g of Hydrogen 3.44 g of Oxygen reacts withX g of Hydrogen X = ( 4 * 3.44 ) / 32     = 0.43 g So, 5.65 - 0.43 = 5.22 g of H2 remain unreacted 32 g of Oxygen produces 36 g of water 3.44 g of  Oxygen produces Y g of water Y = ( 36 * 3.44 ) / 32     = 3.87 g of water B. Which is the limiting reactant? Since the H2 mass is than the required amount , it is thelimiting reactant. C. which is the excess reactant? D. If 2.50 g of H2O is the actual yeild,what is the percentyeild of the reaction? % yield = (actual yield / Theoritical yield ) * 100            = ( 2.5 / 3.87 ) * 100            = 64.6 %

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