Question

If you were to start with 10grams of both reactants, how much water would be produced ? 2H2 +O2 yields 2H2O

Answer 1

`~~11 " g " H_2O` produced

Explanation:

Given: `2 H_2 + O_2 -> 2 H_2O; " "10 g H_2; " "10 g O_2`

Calculate molar masses:

`H_2: " "2(1.008) = 2.016 "g/(mol)"`

`O_2: " "2(15.999) = 31.998 "g/(mol)"`

`H_2O: " "2(1.008) +15.999 = 18.015 "g/(mol)"`

The equation is balanced.

Find the limiting reagent:

`10 g H_2 xx (1 mol H_2)/(2.016 g H_2) xx (2 mol H_2O)/(2 mol H_2) xx (18.015 g H_2O)/(1 mol H_2O)`

`= 89.36 g " "H_2O`

`10 g O_2 xx (1 mol O_2)/(31.998 g O_2) xx (2 mol H_2O)/(1 mol O_2) xx (18.015 g H_2O)/(1 mol H_2O)`

`= 11.26 g " "H_2O`

`O_2` is the limiting reagent.

This means there will only be `~~ 11 g " "H_2O` produced.