Cfrgtkky

If I have $1$ mol of Cheerios, and I dump them on Earth, and they form even layers, how thick are all the layers (if there are any) after all are done? The height/thickness of each Cheerio is $.8$ cm, it’s radius is $1.27$ cm, the Earth’s radius is $6.371 times 10^6$ m. Assume each Cheerio has the same dimensions as listed above, and that the Earth is a perfect sphere (no bumps or grooves).

First thing I did was find the surface area of Earth, which is $5101times10^{14} m^2$. Then, assuming the base of the Cheerio is a circle, I found the area of the smallest box it could fit in, which is $6.45times10^{-14} m^2$. After that, I divided the surface area of the Earth by the area of the base of each Cheerio to see how many I need to layer it once. That number is $7.91times10^{20}$ Cheerios. To find how many layers I can have before I have $0$ Cheerios I set up the equation $(6.022times10^{23})-(7.91times10^{20}x)=0$, where $x$ is how many layers I can have. Because this has to be an integer, I get $761$ layers, meaning $6.09 m$ is the thickness of the layers.

I get the surface are of the Earth to be $4pi times 6.371 times 10 ^{6} = 5.101 times 10 ^{14}$. Then to simplify the fitting together of Cheerios, treat them as squares of side $1.27 times 2 = 2.54cm.$ The surface area of one Cheerio is therefore $(2.54 cm)^2 = 6.45cm^2 = 6.45 times 10 ^{-4} m^2$.

You can then fit $5.101 times 10 ^{14} div 6.45 times 10 ^{-4} = 7.91 times 10 ^{17}$ Cheerios in one layer. Notice that this is a factor of $10^3$ away from your estimation at this point. Then do the final calculation to arrive at a slightly adjusted answer.

It is worth noting that circles can be packed more efficiently than assuming them to be squares, and it may be worth considering this.

You can have lots of fun with a problem like this, depending on how you want to model it.

First of all, these are ridiculously large Cheerios. The real ones are about half that size, maybe a little less. But let's just use the original data.

Are we randomly dumping the Cheerios? If so, you probably have larger gaps between your "squares" and you will have overestimated the number of Cheerios per layer.
You could use probability to estimate the gaps and get a higher pile of Cheerios in the end.
Or maybe then try to model what happens if we shake the Earth gently in order to ensure that its "contents" will "settle in shipping". That could get complicated.

Or are we laying the Cheerios in semi-regular patterns to get as many as we can in each layer? If so, you want to choose an arrangement that puts as many Cheerios as you can in each square meter. Lining them up along a square grid gives you some number of Cheerios, but if you shift every other row $1.27 mathrm{cm}$ to the right you can put the rows closer together and fit a few more rows in the same square meter.
You can fit about $15.5%$ more Cheerios onto a single square meter this way.

In the following, however, I assume the square grid you assumed.

It appears that you omitted the decimal point when you wrote the area of the Earth as
$5101times10^{14} mathrm m^2.$
It should be approximately $5.101times10^{14} mathrm m^2$ instead.
And then you gave the area of the square as $6.45times10^{-14} mathrm m^2,$
whereas it's actually about $6.45times10^{-4} mathrm m^2.$
I'm guessing you used the correct value for the square but the wrong value for the sphere in your calculation, because
$$frac{5101times10^{14}}{6.45times10^{-4}} approx 7.91times 10^{20},$$
just as you wrote. But if you write the numerator correctly as $5.101times10^{14},$
the result is only $7.91times 10^{17}.$
Assuming all layers are the same, we get about $761000$ layers
(the quotient comes out to $761315$ rounded up to an integer, but it makes no sense to get $6$ significant digits when we dropped everything after the third digit in the input to this last calculation).
So now we have a thickness $6090 mathrm m.$

At this thickness you almost have to start worrying about the fact that the outer shell has a larger area than the inner shell and can fit more Cheerios.
But even better, let's consider the weight of a pile of Cheerios more than six kilometers deep.
I would expect all the Cheerios in most of the lower layers to be crushed to dust by the weight of the layers above. Once they're crushed, the layers get a lot thinner because the Cheerios dust can fill in all the gaps. (There's also a lot of extra volume inside the uncrushed Cheerios themselves.)
So now if you want to be a bit more realistic you can do some research into the material strength of Cheerios and the density of crushed and uncrushed Cheerios.
I'm sure General Mills has done this research, but they might not be sharing the results.

My result:

The surface of the Cheerios being $pi r^2$, and the surface of the Earth $pi d^2$, assuming a filling factor $f$ (ratio of filled surface over total surface), one layer counts

$$ffrac{d^2}{r^2}$$ of them.

The total height is given by

$$mathcal Nhfrac{r^2}{fd^2}=6.02cdot 10^{23} 0.008 frac{0.0127^2}{dfrac{pisqrt3}6left(dfrac{4cdot10^7}{pi}right)^2}=5283,m.$$

IMO this is not large enough to justify a correction for the Earth curvature, given the uncertainty on the dimensions and the filling factor.

You'll probably need reinforced Cheerios to avoid slight crushing in the lower layers...