Given 2AL + 6HCL → 2ALCL3 + 3H2, how many grams of aluminum do I need to produce 11 L of hydrogen gas at STP?

Chemistry

1 answer:

Setler79 [48]3 years ago

3 0

Mass of Aluminum= 8.829 g

<h3>Further explanation</h3>

Given

Reaction

2Al + 6HCl → 2AlCl₃ + 3H₂

Required

mass of Aluminum

Solution

At STP, 1 mol gas = 22.4 L

For 11 L of Hydrogen :

= 11 : 22.4

= 0.491

From equation, mol ratio Al : H₂ = 2 : 3, so mol Al :

= 2/3 x mol H₂

= 2/3 x 0.491

= 0.327

Mass Aluminum(Ar=27 g/mol) :

= 0.327 mol x 27 g/mol

= 8.829 g

You might be interested in

Convert each quantity to the indicated units. a. 3.01g to cg. b. 6200m to km

miss Akunina [59]

a. 301 cg

b. 6.2 km

Explanation:

a. knowing that 1 gram (g) is equal to 100 centigrams (cg) we devise the following reasoning:

if 1 g is equal to 100 cg

then 3.01 g are equal to X cg

X = (3.01 × 100) / 1 = 301 cg

b. knowing that 1 kilometer (km) is equal to 1000 meters (m) we devise the following reasoning:

if 1 km is equal to 1000 m

then Y km are equal to 6200 m

Y = (6200 × 1) / 1000 = 6.2 km

Learn more about:

converting units of measurement

brainly.com/question/11300981

#learnwithBrainly

5 0

4 years ago

Read the following chemical equations.

lora16 [44]

This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:

Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰

Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻

Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.