Question

Question #c71a5

Answer 1

Mass of oxygen required: `"16 g"`
Mass of water produced: `"18 g"`

Explanation:

The first thing you need to do is write the balanced chemical equation for this synthesis reaction

`color(red)(2)H_(2(g)) + O_(2(g)) -> 2H_2O_((l))`

Notice that you have a `color(red)(2):1` mole ratio between hydrogen gas and oxygen gas. This tells you that, regardless of how many moles of hydrogen react, you'll always need 2 times fewer moles of oxygen in order for the reaction to take place.

Use hydrogen's molar mass to determine how many moles of hydrogen are present in your 2-g sample

`2cancel("g") * ("1 mole "H_2)/(2.02 cancel("g")) = "0.99 moles"` `H_2`

This many moles of hydrogen would need

`0.99cancel("moles"H_2) * ("1 mole "O_2)/(color(red)(2)cancel("moles"H_2)) = "0.495 moles"` `O_2`

At the same time, this many moles of hydrogen would produce

`0.99cancel("moles"H_2) * ("2 moles "H_2O)/(color(red)(2)cancel("moles"H_2)) = "0.99 moles"` `H_2O`

Use the molar masses of oxygen and water, respectively, to determine how many grams of each the reaction will need/produce.

`0.495cancel("moles") * "32.0 g"/(1cancel("mole"O_2)) = "15.8 g"` `O_2`

and

`0.99cancel("moles"H_2O) * "18.02 g"/(1cancel("mole"H_2O)) = "17.8 g"` `H_2O`

Since you only gave one sig fig for the mass of hydrogen, the answers must be rounded to one sig fig. However, I'll leave them rounded to two sig figs

`m_(O_2) = color(green)("16 g")`

`m_(H_2O) = color(green)("18 g")`