Oxidation of zinc produced a #47.5*mL# volume of dihydrogen at #1*atm# and #273.15*K#. What is the molar quantity of zinc that reacted?

We investigate the redox response...

To find the molar quantity, we apply the outdated Ideal Gas Equation.

Using the ideal gas law $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is temperature:

We rearrange the formula to solve for moles ($n$): $n = \frac{{PV}}{{RT}}$

Given:

• $P = 1 \, atm$
• $V = 47.5 \, mL = 0.0475 \, L$
• $R = 0.0821 \, L \cdot atm/mol \cdot K$ (gas constant)
• $T = 273.15 \, K$

Now, plug in the values: $n = \frac{{(1 \, atm) \times (0.0475 \, L)}}{{(0.0821 \, L \cdot atm/mol \cdot K) \times (273.15 \, K)}}$

$n = \frac{{0.0475}}{{22.4147}}$

$n \approx 0.00212 \, moles$

Therefore, approximately 0.00212 moles of zinc reacted.

The molar quantity of zinc that reacted is 0.00181 moles.