In a chemical reaction you collect 1L Hydrogen gas at STP. How many moles of hydrogen gas did you collect?

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In a chemical reaction you collect 1L Hydrogen gas at STP. How many moles of hydrogen gas did you collect?

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Answer 1 · 2018-03-13T21:47:33

$0.0446 mol$ $"0.0446 mol"$ .

Explanation:

In this question, we're given:

$p$ $p$ , or pressure, of $1 atm$ $"1 atm"$ . We know this because it tells us that we're at STP.
$V$ $V$ , or volume, of $1 L$ $"1 L"$ .
$T$ $T$ , or temperature, at $273.15 K$ $"273.15 K"$ . We know this because it tells us that we're at STP.

We need to find $n$ $n$ , or number of moles.
The Ideal Gas Equation relates all of these variables!:

$p V = n R T$ $pV = nRT$

We can rearrange this to only have $n$ $n$ on one side:

$n = \frac{p V}{R T}$ $n = (pV)/(RT)$

We know the value of $R$ $R$ , or the Universal Gas Constant, to be $0.08206 L atm/K mol$ $"0.08206 L atm/K mol"$ because the units for our given values are in $L$ $"L"$ , $atm$ $"atm"$ , and $K$ $"K"$ .
Now, we can just plug in the variables and solve:

$n = \frac{1 atm \times 1 L}{0.08206 L atm/K mol \times 273.15 K}$ $n = ("1 atm" xx "1 L")/("0.08206 L atm/K mol" xx "273.15 K")$
$n = (1 cancel("atm") xx 1 cancel("L"))/(0.08206 cancel("L") cancel("atm")"/"cancel("K") "mol" xx 273.15 cancel("K"))$
$n = "0.0446 mol"$

We can verify that our answer is correct by plugging everything into the Ideal Gas Equation again and seeing if $pV$ is really equal to $nRT$ :

$pV = nRT$
$"1 atm" xx "1 L" = "0.0446 mol" xx "0.08206 L atm/K mol" xx "273.15 K"$
$1 = 1$

Yay! :)

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In a chemical reaction you collect 1L Hydrogen gas at STP. How many moles of hydrogen gas did you collect?

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