What mass in grams of hydrogen is produced by the reaction of 4.73g of Magnesium with 1.83g of water?

Question

What mass in grams of hydrogen is produced by the reaction of 4.73g of Magnesium with 1.83g of water?

Mg (s) 2H2O (l) --> Mg(OH)2 (s) H2 (g)

THANK YOU!!!

Answer 1

nope,

the above answer at the point:

.....(1 mole H2 / 2 moles H2O) = 0.051g H2

has calculated the # of moles of H2 ,... grams will need another step where  you multiply by 2.0156 grams of H2 per mole,... in order to get the grams of H2 produced = 0.103 g H2

======================================...

Mg (s) 2H2O (l) --> Mg(OH)2 (s) H2 (g)

2 moles of water @ 18.02g/mol = 36.04 grams of water

1 mol of Mg has a molar mass of 24.30 grams

they ratio that reacts is 36.04 grams of H2O / 24.30 grams Mg = 1.48

the ratio that they used was 1.83 g H2O / 4.73 g Mg = 0.387

they have seriously cut back on the amount of water, water is your limiting reagent:

---------------

so:

What mass in grams of hydrogen is produced by the reaction with 1.83g of wate, knowing that the balanced equation predicts 1 mole of Hydrogen @ 2.02g/mol will be produced from 36.04 grams of water:

1.83 grams H2O @ 2.02 grams of H2 / 36.04 grams of water= 0.10257 grams of hydrogen

your answer is: 0.103 grams of H2

Answer 2

Mg (s) +  2H2O (l) --> Mg(OH)2(s)+  H2 (g)

4.73g Mg x (1 mole Mg / 24.305g Mg) x (1 mole H2 / 1 mole Mg) = 0.196g H2

1.83g H2O x (1 mole H2O / 18.001g H2O) x (1 mole H2 / 2 moles H2O) = 0.051g H2

H2O is your limiting reagent and 0.051g H2 is produced