Figuring out the theoretical mass of reactants
Try to answer to these questions, don't cheat, check the answer passing over the yellow box.
I think this software is confusing you always start from how much reactants you have and then calculate how much products you will have, suppose you have $\pu{100g}$ of oxygen and $\pu{44g}$ of propane.
How many mole of propane and oxygen ($\ce{O2}$) do you have?

 Oxygen moles$=\frac{\pu{100g}}{\pu{32g/mol}}\sim \pu{3mol}$,  Propane moles$=\frac{\pu{44g}}{\pu{44.096g/mol}}\sim \pu{1 mol}$

If this is your balanced equation:
$$\ce{C3H8 + 5O2 -&gt; 3CO2 + 4H2O}$$
How many oxygen molecules do you need to react with one molecule of propane?

 You need 5 oxygen molecules for every propane molecule. So you need 5 moles of oxygen for every mole of propane.

Which is the ratio?

 $\frac{1}{5}$ if you have $x$ moles of oxygen you need $\frac{x}{5}$ moles of propane. e.g. if you have 5 moles of oxygen you need? 1 mole of propane.

But you have only 3 moles of oxygen to get the products, oxygen is a limiting reagent! So how many moles of propane will react:

 $$\pu{\frac{3}{5}mol} =\pu{0.6 mol} $$

So how many moles of the products will you have?

 $3\times0.6$ moles of $\ce{CO2}$ and $4\times0.6$ moles of water

And so how many grams?

 for $\ce{CO2}$, $\pu{1.8mol}\times \pu{44g/mol}$ for water, $\pu{2.4mol}\times \pu{18g/mol}$