How many moles of water are produced when 10 moles of hydrogen gas reacts with an excess of oxygen? (whole number)

12 moles H

2

O

Explanation:

Your tools of choice for stoichiometry problems will always be the mole ratios that exist between the chemical species that take part in the reaction.

As you know, the stoichiometric coefficients attributed to each compound in the balanced chemical equation can be thought of as moles of reactants needed or moles of products formed in the reaction.

In your case, the balanced chemical equation for this synthesis reaction looks like this

2

H

2(g]

+

O

2(g]

→

2

H

2

O

(l]]

Notice that the reaction requires  

2

moles of hydrogen gas and  

1

mole of oxygen gas to produce  

2

moles of water.

This tells you that the reaction produces twice as many moles of water as you have moles of oxygen gas that take part in the reaction.

You know that your reaction uses  

6.0

moles of oxygen. Assuming that hydrogen gas is not a limiting reagent, you can say that the reaction will produce

6.0

moles O

2

⋅

2

moles H

2

O

1

moles O

2

=

12 moles H

2

O

Explanation:

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%