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Gnom [1K]
3 years ago
10

When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coeffici

ents are:
aluminum oxide (s) + sulfuric acid (aq) ----> aluminum sulfate (aq) + water (l)

Does a reaction occur when aqueous solutions of ammonium carbonate and chromium(III) nitrate are combined?
Chemistry
1 answer:
maxonik [38]3 years ago
7 0

Answer:

1. Al2O3 + 3H2SO4 —> Al2(SO4)3 + 3H2O

The coefficients are: 1, 3, 1, 3

2. Yes, reaction occurs when ammonium carbonate and chromium(III) nitrate are combined.

The balanced equation for the reaction is given below:

3(NH4)2CO3 + 2Cr(NO3)3 —> 6NH4NO3 + Cr2(CO3)3

The coefficients are: 3, 2, 6, 1

Explanation:

1. The word equation is given below:

aluminum oxide(s) + sulfuric acid(aq) -> aluminum sulfate(aq) + water (l)

The above can be written as:

Al2O3 + H2SO4 —> Al2(SO4)3 + H2O

Now we can balance the equation as follow:

There are 3 atoms of SO4 on the right side and 1 atom on the left. It can be balance by putting 3 in front of H2SO4 as shown below:

Al2O3 + 3H2SO4 —> Al2(SO4)3 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 3 in front of H2O as shown below:

Al2O3 + 3H2SO4 —> Al2(SO4)3 + 3H2O

Now the equation is balanced

The coefficients are: 1, 3, 1, 3

2. Yes, reaction occurs when ammonium carbonate and chromium(III) nitrate are combined.

The equation for the reaction is given below:

(NH4)2CO3 + Cr(NO3)3 —> NH4NO3 + Cr2(CO3)3

We can balance the equation above as follow:

There are 2 atoms of Cr on the right side and 1 atom on the left. It can be balance by putting 2 in front of Cr(NO3)3 as shown below:

(NH4)2CO3 + 2Cr(NO3)3 —> NH4NO3 + Cr2(CO3)3

Now, there are 6 atoms of NO3 on the left side and 1 on the right side. It can be balance by putting 6 in front of NH4NO3 as shown below:

(NH4)2CO3 + 2Cr(NO3)3 —> 6NH4NO3 + Cr2(CO3)3

Now, there are 2 atoms of NH4 on the left side and 6 atoms on the right. It can be balance by putting 3 in front of (NH4)2CO3 as shown below:

3(NH4)2CO3 + 2Cr(NO3)3 —> 6NH4NO3 + Cr2(CO3)3

Now the equation is balanced.

The coefficients are: 3, 2, 6, 1

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<u>4, 7, 4, 6</u>

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<u>12 moles</u>

Explanation:

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

__↑______↑

8.00 mol | 14.00 mol

________________

NH_{3}(g) + O_{2}(g) \: → NO_{2} + H_{2}O(g)

You can turn this into a system of variables which are solvable.

To do this, create variables for the coefficients of each compound in the reaction respectively.

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Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.

a corresponds to the coefficient of the first compound, b corresponds to the coefficient of the second compound, c corresponds to the coefficient of the third compound, and d corresponds to the coefficient of the fourth compound.

(Reactant = Product)

Reactant: 1a [N] Product: 1c.

Reactant: 3a [H] Product: 2d.

Reactant: 2b [O] Product: 2c + 1d.

Thus the system is:

1a = 1c

3a = 2d

2b = 2c + 1d.

Then just use the substitution methods to solve.

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