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Leno4ka [110]
3 years ago
6

If 8.82 grams of Aluminum reacts, how many grams of AlCl3 will be produced?

Chemistry
1 answer:
loris [4]3 years ago
8 0

Hey there! :

Molar mass:

Al = 26.9815 g/mol

AlCl₃ = 133.3405 g/mol

Reaction stochemistry ,calculates the mass of AlCl3 produced

2 Al + 3 Cl₂ =  2 AlCl₃

2 * 26.9815 g Al ---------------------  2 *  133.3405 g AlCl₃

8.82 g Al -------------------------------  ( mass of AlC₃ )

mass of AlCl₃  =  ( 8.82 * 2 * 133.3405 ) /  ( 2 * 26.9815 )

mass of AlCl₃ = 2352.12642 / 53.963

<u>mass of AlCl₃ = 43.58 g of AlCl₃</u>

Hope this helps!

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We'll begin by calculating the number of mole of Pb(NO₃)₂ in the solution.

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Mole = Molarity x Volume

Mole of Pb(NO₃)₂ = 0.25 × 0.01

Mole of Pb(NO₃)₂ = 0.0025 mole

Next, we shall determine the mole of LiBr required to react with 0.0025 mole of Pb(NO₃)₂

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From the balanced equation above,

1 mole of Pb(NO₃)₂ reacted with 2 mole of LiBr.

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0.0025 mole of Pb(NO₃)₂ will react with = 2 × 0.0025 = 0.005 mole of LiBr

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Molarity of LiBr = 0.005 / 0.025

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