Question

Question #91e81

Answer 1

`"0.27 g H"_2`

Explanation:

The first thing to do here is write a balanced chemical equation for this single replacement reaction

`color(red)(2)"Al"_text((s]) + 6"HCl"_text((aq]) -> 2"AlCl"_text(3(aq]) + color(blue)(3)"H"_text(2(g]) uarr`

You're interested in finding out how much hydrogen gas will be produced by your sample of aluminium, so the `color(red)(2):color(blue)(3)` mole ratio that exists between the two chemical species is important here,

Now, focus on using the **percent composition of the alloy to figure out how much aluminium you have in your `"2.6-g"` sample.

Since you know that the alloy is `93.7%` aluminium by mass, which is equivalent to saying that out of every `"100 g"` of alloy, `"93.7 g"` will be aluminium, you will have

`2.6 color(red)(cancel(color(black)("g alloy"))) * "93.7 g Al"/(100color(red)(cancel(color(black)("g alloy")))) = "2.436 g Al"`

Next, use aluminium's molar mass to find how many moles of aluminium you have in that sample

`2.436 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "0.09029 moles Al"`

Since you know that every `color(red)(2)` moles of aluminium will produce `color(blue)(3)` moles of hydrogen gas, you will have

`0.09029color(red)(cancel(color(black)("moles Al"))) * (color(blue)(3)" moles H"_2)/(color(red)(2)color(red)(cancel(color(black)("moles Al")))) = "0.1354 moles H"_2`

Finally, use the molar mass of hydrogen gas to find you many grams would contain this many moles

`0.1354 color(red)(cancel(color(black)("moles H"_2))) * "2.0159 g"/(1color(red)(cancel(color(black)("mole H"_2)))) = color(green)("0.27 g H"_2)`

The answer is rounded to two sig figs, the number of sig figs you have for the mass of the alloy.