If the reaction yield is 94.4%, what mass in grams of hydrogen is produced by the reaction of 4.73 g of magnesium with 1.83 g of water? mg(s)+2h2o(l)→mg(oh)2(s)+h2(g)

First, it is important to list the molar mass of the relevant substances.

Molar mass of magnesium = 24.305 g/mol
Molar mass of water = 18.0153 g/mol
Molar mass of H2 = 2.0159 g/mol

Second, we need to determine the limiting reactant for the chemical reaction. We take 4.73 g of Mg and determine the stoichiometric amount of water needed for it to be completely consumed. This is shown in the following equation:

4.73 g Mg x mol Mg/24.305 g x 2 mol H2O/1 mol Mg x 18.0153 g/mol H2O = 7.0119 g H2O

Thus, we have determined that 7.0119 g H2O is needed to completely react 4.73 g Mg. The given amount of 1.83 g H2O is insufficient which then indicates that water is the limiting reactant and should be the basis of our calculations. 

Next, given 1.83 g H20, we calculate the theoretical yield of hydrogen gas using stoichiometry. The equation is then:

1.83 g H2O x mol H20/18.0153 g x 1 mol H2/2 mol H2O x 2.0159 g/mol H2 = 0.1024 g H2

However, the reaction yield was given to be 94.4%. The reaction yield is given by the formula percent yield = actual yield/ theoretical yield x 100%. Thus, the actual yield of hydrogen gas can be determined using the formula. 

Actual yield of H2 = 0.94*0.1024 g H2

Thus, the amount of hydrogen gas produced is 0.0963 g.   

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