2H2(g) + 02(g) ----> 2H20(g) Calculate the mass (in grams) of water vapor produced when 56.0 g of pure?

Question

Calculate the mass (in grams) of water vapor produced when 56.0 g of pure hydrogen gas burns in air.
Explain answer

Answer 1

2 H2(g) + 02(g) ---> 2 H20 (g)
First, we need to know the number of moles of H2 used as reactant.
56.0 g H2 x 1 mol H2/2.016 g = 27.8 mol H2
So ,the ratio between H2 and O2 = 2 : 1
The number of oxygen needed in this reaction = 1/2 x 27.8 = 13.9 moles
Actually, we don't have to know how many moles of 02 needed in this reaction.

The number of moles of water produced = the number of moles of H2 = 27.8 moles of water x 18.016 g/mol H20 = 500.84 g H20
So that's the final answer : the mass of water produced = 500.8 g

Answer 2

note that once 2 moles of hydrogen are used, 2 moles of water are produced. 56g/2.0.5 =27.8 mole of Hydrogen used 27.8 moles of water will be produced. 27.8mol x 18g/mol = 500 g or 501g rounding up.

Answer 3

Notice that when 2 moles of hydrogen are used, 2 moles of water are produced.
56g/2.015 =27.8 mole of Hydrogen used
27.8 moles of water will be produced.
27.8mol x 18g/mol = 500 g or 501g rounding up.