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Sedbober [7]
3 years ago
9

CaSO4 + AlBr3 > CaBr2 +Al2(SO4)3 What will the coefficient of CaBr2 when the equation is balanced using the smallest possible

whole number coefficient
Chemistry
2 answers:
Deffense [45]3 years ago
8 0

Answer:

3 will be the correct coefficient of CaBr2

Explanation:

In balancing a chemical equation, numbers should be assigned to both reactants and products as a numerical coefficients until all atoms of elements in both sides of the equation count equal.

The balanced equation of the reaction will be:

3CaSO4 + 2AlBr3 ==> 3CaBr2 + Al2(SO4)3

Looking at the unbalanced equation in the question, in the product Al2(SO4)3 there are 3 SO4 group. This will warrant putting 3 behind CaSO4 in order to balance the atoms of SO4 group. That operation will automatically put the number of Ca atoms in CaSO4 to be 3 therefore making CaBr2 to have 3 coefficient as in the balanced equation. This is to balance the number of Ca atoms in both sides to be 3.

Umnica [9.8K]3 years ago
8 0

Answer:

The coefficient of CaBr2 is 3

Explanation:

Step 1: The unbalanced equation

CaSO4 + AlBr3 → CaBr2 +Al2(SO4)3

Step 2: Balancing the equation

CaSO4 + AlBr3 → CaBr2 +Al2(SO4)3

On the left side we 1x SO4 (in CaSO4), on the right side we have 3x SO4 (in Al2(SO4)3. To balance the amount of SO4 on both sides, wehave to multiply CaSO4 by 3.

3 CaSO4 + AlBr3 → CaBr2 +Al2(SO4)3

On the left side we have 1x Al (AlBr3) and on the right side we have 2x Al (in Al2(SO4)3. To balance the amount of Al on both sides,we have to multiply AlBr3 on the left side, by 2.

3 CaSO4 + 2AlBr3 → CaBr2 +Al2(SO4)3

On the left side we have 3x Ca (in CaSO4), on the right side we have 1x Ca (in CaBr2). To balance the amount of Ca on both sides, we have to multiply CaBr2 on the right side, by 3. Now the equation is balanced.

3 CaSO4 + 2AlBr3 → 3CaBr2 +Al2(SO4)3

The coefficient of CaBr2 is 3

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Answer:- 14.9 M

Solution:- Given commercial sample of ammonia is 28% by mass. Let's say we have 100 grams of the sample. Then mass of ammonia would be 28 grams.

Density of the solution is given as 0.90 grams per mL.

From the mass and density we could calculate the volume of the solution as:

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Let's convert the volume from mL to L as molarity is moles of solute per liter of solution.

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Now, we convert grams of ammonia to moles on dividing the grams by molar mass. Molar mass of ammonia is 17 gram per mole.

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molarity=\frac{1.65mole}{0.111L}

= 14.9 M

So, the molarity of the given commercial sample of ammonia is 14.9 M.

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