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inn [45]
3 years ago
3

In some fireworks there is a reaction between powdered aluminium and powdered barium nitrate in which heat is evolved and an unre

active gas is produced.
What is the equation for this reaction?

A. 2Al + Ba(NO3)2 -> Al2O3 + BaO + 2NO
B. 4Al + 4Ba(NO3)2 -> 2Al2O3 + 4 Ba(NO2)2 + O2
C. 10Al + 3 Ba(NO3)2 -> 5Al2O3 + 3BaO + 3N2
D. 10Al + 18Ba(NO3)2 -> 10Al(NO3)3 +18BaO +3N2
Chemistry
2 answers:
Dmitriy789 [7]3 years ago
7 0
I think the correct answer from the choices listed above is option A. The correct equation would be:

<span> 2Al + 3Ba(NO3)2 -> Al2O3 + 3BaO + 6NO2
</span>
The products would be aluminum nitrate, barium oxide and nitrogen dioxide.
wariber [46]3 years ago
7 0

<u>Answer:</u> The correct answer is Option 3.

<u>Explanation:</u>

When aluminium in solid state reacts with powdered barium nitrate, it results in the formation of number of products.

The chemical reaction between aluminium and barium nitrate follows the equation:

10Al+3Ba(NO_3)_2\rightarrow 5Al_2O_3+3BaO+3N_2

By Stoichiometry of the reaction:

10 moles of aluminium reacts with 3 moles of barium nitrate to produce 5 moles of aluminium oxide, 3 moles of barium oxide and 3 moles of nitrogen gas.

Nitrogen gas evolved is the unreactive gas.

Hence, the correct answer is Option 3.

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Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis
almond37 [142]

Answer: a) BaCl_2

b)  BiO_3H_3

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba =\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = \frac{0.48}{0.48}=1

For O =\frac{0.96}{0.48}=2

The ratio of Ba: Cl= 1:2

Hence the empirical formula is BaCl_2

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi =\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles

Moles of O= \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= \frac{0.38}{0.38}=1

For O =\frac{1.15}{0.38}=3

For H=\frac{1.16}{0.38}=3

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is BiO_3H_3

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