stoichiometry

Question

What mass of water can be formed by igniting 27.5 g of hydrogen gas and 175 g of oxygen gas?

Answer 1

As in any chemical reaction, one must first write a correctly balanced equation.

2H₂(g) + O₂(g) ==> 2H₂O(g)

 

Next, determine which reactant is present in limiting supply:

moles H₂ present = 27.5 g x 1 mole/2 g = 13.75 moles H₂

moles O₂ present = 175 g x 1 mole/32 g = 5.47 moles O₂

Since, from the balanced equation, it takes 2 moles H₂ for each 1 mole O₂, O₂ will be limiting since there is enough H₂.

 

Now, based on O₂ being limiting, determine moles of H₂O formed, and then convert that to grams of H₂O formed:

5.47 moles O₂ x 2 moles H₂O/mole O₂ = 10.94 moles H₂O

10.94 moles H₂O x 18 g/mole = 197 g H₂O formed