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3 years ago

Balance the following equation on scrap paper: AIF3+Li2O → Al2O3 + LiF

Chemistry

1 answer:

Oxana [17]3 years ago

8 0

Answer:

2AlF₃ + 3Li₂O —> Al₂O₃ + 6LiF

Explanation:

AlF₃ + Li₂O —> Al₂O₃ + LiF

The above equation can be balanced as follow:

AlF₃ + Li₂O —> Al₂O₃ + LiF

There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by writing 2 before AlF₃ as shown below:

2AlF₃ + Li₂O —> Al₂O₃ + LiF

There are 6 atoms of F on the left side and 1 atom on the right side. It can be balance by writing 6 before LiF as shown below:

2AlF₃ + Li₂O —> Al₂O₃ + 6LiF

There are 2 atoms of Li on the left side and 6 atoms on the right side. It can be balance by writing 3 before Li₂O as shown below:

2AlF₃ + 3Li₂O —> Al₂O₃ + 6LiF

Thus, the equation is balanced..!

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Read 2 more answers

Calculate the mass of ethyl alcohol required to prepare 540 grams of C4H6, if the reaction follows the scheme:

kkurt [141]

Answer:

Explanation:

2C₂H₅OH = C₄H₆ + 2H₂O + H₂

2 mole 1 mole

molecular weight of ethyl alcohol

mol weight of C₂H₅OH = 46 gm

mol weight of C₄H₆ 54 gm

540 gm of C₄H₆ = 10 mole

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3 0

3 years ago

Consider the reaction 2Al + 6HBr → 2AlBr3 + 3H2. If 8 moles of Al react with 8 moles of HBr, what is the limiting reactant?

TiliK225 [7]

Answer:- HBr is limiting reactant.

Solution:- The given balanced equation is:

$2Al+6HBr\rightarrow 2AlBr_3+3H_2$

From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.

The calculations could be shown as:

$8molAl(\frac{6molHBr}{2molAl})$

= 24 mol HBr

From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.

7 0

3 years ago