Question

Question #f784e

Answer 1

`"88.1 mL"`

Explanation:

Indeed, you must write the balanced chemical equation for this single replacement reaction.

Aluminium will react with sulfuric acid, `"H"_2"SO"_4`, to form aluminium sulfate, `"Al"_2("SO"_4)_3`, and hydrogen gas, `"H"_2`, which will bubble out of solution.

`2"Al"_text((s]) + color(purple)(3)"H"_2"SO"_text(4(aq]) -> "Al"_2("SO"_4)_text(3(aq]) + color(blue)(3)"H"_text(2(g]) uarr`

Now, notice that you have a `color(purple)(3) : color(blue)(3)` mole ratio between sulfuric acid and hydrogen gas.

This tells you that, regardless of how many moles of sulfuric acid react, the reaction will produce the same number of moles of hydrogen gas.

In other words, if your reaction produced a certain number of moles of hydrogen gas, then the same number of moles of sulfuric acid took part in the reaction.

Use the ideal gas law equation to determine how many moles of hydrogen gas were produced.

`color(blue)(PV = nRT implies n = (PV)/(RT))`

Remember to convert the volume from milliliters to liters and the pressure from torr to atm!

Also, I will assume that the volume of the gas is equal to `"3255 mL"`, not `"3.255 mL"`!

`n = (975/760 color(red)(cancel(color(black)("atm"))) * 3255 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 385 color(red)(cancel(color(black)("K")))) = "0.1321 moles H"_2`

So, if the reaction produced `0.1321` moles of hydrogen, then you can conclude that the reaction used up `0.1321` moles of sulfuric acid!

`0.1321 color(red)(cancel(color(black)("moles H"_2))) * (color(purple)(3)" moles H"_2"SO"_4)/(color(blue)(3)color(red)(cancel(color(black)("moles H"_2)))) = "0.1321 moles H"_2"SO"_4`

Now all you have to do is use the molarity of sulfuric acid solution to determine what volume would contain this many moles of sulfuric acid

`color(blue)(c = n/V implies V = n/c)`

`V = "0.1321 moles"/(1.500 color(red)(cancel(color(black)("moles")))/"L") = "0.08807 L"`

Rounded to three sig figs and expressed in milliliters, the answer will be

`V = color(green)("88.1 mL")`