Question #61ae6

The idea here is that you need to use the molar masses of the two elements to convert the mass ratio to a mole ratio.

You know that you have

#M_ ("M H"_2) ~~ "2 g mol"^(-1)#

#M_ ("M O"_2) ~~ "32 g mol"^(-1)#

Right from the start, you can say that in a mixture that contains equal masses of hydrogen gas and oxygen gas, the two gases will be in a #16:1# mole ratio, i.e. you will have #16# times more moles of hydrogen gas than of oxygen gas.

Now, you know that your mixture contains hydrogen gas and oxygen gas in a #1:4# mass ratio.

This means that you have #4# times as many grams of oxygen gas than of hydrogen gas, which implies that you have #4# times as many moles of oxygen gas than you would have if the two gases had equal masses in the mixture.

This implies that in this case, the #16:1# mole ratio will be equal to #16:4 = 4:1#.

Therefore, you can say that the mole ratio that exists between the two gases is equal to

#"H"_2/"O"_2 =4/1#

To double-check the logic, take #m# to be the mass of hydrogen gas present in the sample #-># the mass of oxygen gas will be #4 * m#.

The number of moles of hydrogen gas will be

#m color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2color(red)(cancel(color(black)("g")))) = (m/2)# #"moles H"_2#

The number of moles of oxygen gas will be

#4m color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32color(red)(cancel(color(black)("g")))) = (m/8)# #"moles O"_2#

The mole ratio that exists between the two gases will be equal to

#"H"_2/"O"_2 = (color(red)(cancel(color(black)("m")))/2 color(red)(cancel(color(black)("moles"))))/(color(red)(cancel(color(black)("m")))/8 color(red)(cancel(color(black)("moles")))) = 1/2 * 8/1 = 4/1#