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.