Question

Suppose you ran this same reaction (using the balloon setup as seen in the video) on your own with two different flasks. In Flask A, you reacted 5.10 g Mg with 0.447 mol HCl. In Flask B, you reacted 24.21 g Mg with 0.998 mol HCl. Which ballon will inflate the most? Explain (in detail) why, showing all work.

Answer 1

the set-up that produces the most gas will inflate the most.

Answer 2

How do I prove that?

Answer 3

stoichiometry

Answer 4

write and balance the equation for the process and use the mass (in terms of moles) to find the amount of \(H_2\) produced.

Answer 5

Lol I don't exactly know how to do that yet.

Answer 6

Okay, first write the equation for the process.
Convert the mass given to moles. Now find the limiting reactant by dividing the moles of each reactant by it's stoichiometric coefficient in the reaction (the number in front of it).
Which ever has less moles after the division is the limiting reactant.
Now, use the moles of the limiting reactant to find the moles of hydrogen gas produced.

Set up a ratio using the species of interest, like so:

e.g. for a general reaction:

\(\color{red}{a}A + \color{blue}{b}B\) \(\rightleftharpoons\) \( \color{green}{c}C\)

where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients ,

\(\dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\)

From here you can isolate what you need.
For example: if you have 2 moles of B, how many moles of C can you produce?

solve algebraically: \(\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)
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To convert mass to moles, use the relationship:

\(n=\dfrac{m}{M}\)

where, M=molar mass, m=mass, and n= moles.