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kari74 [83]
3 years ago
7

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

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In the following reaction, how many liters of o2 will produce 43.62 liters of co2 at stp? c3h8 5 o2 3 co2 4 h2o
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The balanced chemical reaction:

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We are given the amount of the carbon dioxide to be produced. This will be the starting point of our calculations.

<span>43.62 L CO2 ( 1 mol CO2 / 22.4 L CO2 ) (5 mol O2 / 3 mol CO2 ) ( 22.4 L O2 / 1 mol O2) = 72.7 L O2</span>



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Read 2 more answers
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