Question

How many grams of Al were reacted with excess HCI if 5.36 L of hydrogen gas were collected at STP in the following reaction ? 2Al(s) + 6HCl(aq) -> 2AlC*l_{3}; (aq) + 3H_{2}(g)

Answer 1

In this reaction, 2 moles of Al react with 6 moles of HCl to produce 3 moles of hydrogen gas at STP.

First, let's calculate the number of moles of hydrogen gas produced:

Using the ideal gas law, PV = nRT, we can find the number of moles of hydrogen gas produced:

n = PV/RT

where P is the pressure (1 atm at STP), V is the volume of hydrogen gas collected (5.36 L), R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature (273 K at STP).

n = (1 atm)(5.36 L)/(0.0821 L·atm/mol·K)(273 K) = 0.232 mol H2

Since 2 moles of Al react with 3 moles of H2, we can find the number of moles of Al that reacted:

mol Al = (2/3) mol H2 = (2/3) (0.232 mol) = 0.155 mol Al

Finally, we can convert the number of moles of Al to grams using the molar mass of Al:

mass Al = mol Al × molar mass Al = 0.155 mol × 26.98 g/mol = 4.17 g Al

Therefore, 4.17 grams of Al were reacted with excess HCl

Answer 2

11.24 g of Al were reacted with excess HCl.

Step-by-step explanation:

To determine the number of grams of aluminum (Al) that were reacted with hydrochloric acid (HCl), we need to use the balanced chemical equation and the Ideal Gas Law.

The balanced chemical equation for the reaction is:

2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)

From this equation, we can see that for every 2 moles of Al that react, 3 moles of H2 are produced.

Given that 5.36 L of hydrogen gas were collected at STP (standard temperature and pressure), which is 0°C and 1 atm, we can use the Ideal Gas Law to find the number of moles of H2 produced:

PV = nRT

Where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin.

At STP, the pressure is 1 atm and the volume is 5.36 L, so we can plug in the values:

1 atm * 5.36 L = n * 0.0821 L * atm / mole * (273.15 + 0) K

Rearranging and solving for n, we get:

n = (1 atm * 5.36 L) / (0.0821 L * atm / mole * (273.15 + 0) K) = 0.625 moles

Since the reaction produces 3 moles of H2 for every 2 moles of Al, we can divide the number of moles of H2 by the ratio of moles of Al to moles of H2:

0.625 moles / (3 moles H2 / 2 moles Al) = 0.417 moles Al

Finally, we can convert moles to grams using the atomic weight of Al, which is 26.98 g/mol:

0.417 moles Al * (26.98 g Al / 1 mole Al) = 11.24 g Al

So, 11.24 g of Al were reacted with excess HCl.