Stoichiometry?

Question

What is the maximum mass of hydrogen produced by the reaction of 4.73 g magnesium with 1.83 g of water?

Mg + 2H2O ---> Mg(OH)2 + H2

Answer 1

Very simple, my friend.

You must first find the limiting reactant. You do this by figuring out first how many moles of each reactant that you have. 4.73 g of magnesium and 1.83 g of water is what the problem tells you. To convert to moles, you must divide by the molar mass.

Mg's molar mass is 24.3, so (4.73g/24.3g/mol) = .195mol

H20's molar mass is (1+1+16) = 18, so (1.83g/18g/mol)= .102mol

Since the ratio of moles of water to magnesium is 2:1, there should be twice as many moles of water as Mg. However, there is less water, so water is the limiting reactant.

If the maximum .102 moles of water are used, half that amount of H2 will be produced since the balanced chemical equation shows a 2:1 ratio of water to hydrogen.

SO, dividing .102 by 2 gives us .051 moles of H2.

The final step is to convert moles of H2 to mass, in grams, which requires multiplication.

.051 moles * 2 g/mol H2 = .102 g H2

DoNe

Answer 2

for each mol of water you put in you have 1 mol of H2 being produced.

the mol of water you put in is 1.83/18 = 0.1
half of that will end up being h2. so you will produce 0.05 mol of h2. the molar mass of h2 is 2. so the total maximum mass or h2 is 0.05 x 2 =0.1

DONE

Answer 3

36 g will need 24 g

So, 1.83 g will take up 1.2 g (approx)

1.2 g of Mg will liberate 0.1 g of H2.