# Aluminum reacts with excess hydrochloric acid to form aqueous aluminum chloride and 46.5 mL of hydrogen gas over water at 27 degrees Celsius and 751 mmHg. How many grams of aluminum reacted? The partial pressure of water at 27 C is 26.8 mmHg.

Jul 8, 2017

$0.0324$ $\text{g Al}$

#### Explanation:

We're asked to find the number of grams of $\text{Al}$ that reacted, given some ${\text{H}}_{2} \left(g\right)$ product characteristics.

Let's first write the chemical equation for this reaction:

$2 {\text{Al"(s) + 6"HCl"(aq) rarr2 "AlCl"_3(aq) + 3"H}}_{2} \left(g\right)$

The total pressure of the gaseous system is given as $751$ $\text{mm Hg}$, and the partial pressure of water vapor is $26.8$ $\text{mm Hg}$ at ${27}^{\text{o""C}}$. The pressure of hydrogen gas is thus

P_"total" = P_ ("H"_2"O") + P_ ("H"_2)

${P}_{{\text{H}}_{2}} = 751$ $\text{mm Hg}$ $- 26.8$ $\text{mm Hg}$ = color(red)(724 color(red)("mm Hg"

This pressure in atmospheres is

724cancel("mm Hg")((1color(white)(l)"atm")/(760cancel("mm Hg"))) = color(red)(0.953 color(red)("atm"

We'll now use the ideal gas equation to find the number of moles of ${\text{H}}_{2}$ formed:

($T = {27}^{\text{o""C}} + 273 = 300$ $\text{K}$)

n = (PV)/(RT) = ((color(red)(0.953)cancel(color(red)("atm")))(0.0465cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(300cancel("K"))) = color(green)(0.00180 color(green)("mol H"_2

(volume converted to liters here)

Using the coefficients of the chemical equation, we'll now find the relative number of moles of $\text{Al}$ that react:

color(green)(0.00180)cancel(color(green)("mol H"_2))((2color(white)(l)"mol Al")/(3cancel("mol H"_2))) = color(purple)(0.00120 color(purple)("mol Al"

Lastly, we'll use the molar mass of aluminum ($26.98$ $\text{g/mol}$) to find the number of grams that reacted:

color(purple)(0.00120)cancel(color(purple)("mol Al"))((26.98color(white)(l)"g Al")/(1cancel("mol Al"))) = color(blue)(0.0324 color(blue)("g Al"

Thus, color(blue)(0.0324 sfcolor(blue)("grams of aluminum" reacted.