Question #c71a5

1 Answer
Jun 16, 2015

Answer:

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")#