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ExtremeBDS [4]
3 years ago
13

Balance this equation and pick the answer with the coefficients that belong there.

Chemistry
1 answer:
geniusboy [140]3 years ago
6 0

Answer:

2,3,6,1

2,3,6,1

Explanation:

The unbalanced reaction expression is given as:

  AlBr₃    +    K₂SO₄  →    KBr  + Al₂(SO₄)₃

We need to balanced this reaction equation. Our approach is a mathematical method where we assign variable a,b,c and d as the coefficients.

 aAlBr₃    +    bK₂SO₄  →    cKBr  + dAl₂(SO₄)₃

Conserving Al;  a  = 2d

                   Br:  3a  = c

                   K:   2b = c

                   S:   b  = 3d

                   O:  4b  = 12d

Let a  = 1, c = 3, d  = \frac{1}{2}  b  = \frac{3}{2}  

 Multiply through by 2 to give;

     a  = 2, b = 3, c = 6 and d  = 1

 2AlBr₃    +    3K₂SO₄  →    6KBr  + Al₂(SO₄)₃

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Solid sodium azide (NaN3) produces solid sodium and nitrogen gas. How many grams of sodium azide are needed to yield a volume of
ch4aika [34]

Answer:

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P*V = n*R*T

In this case, the balanced reaction is:

2 NaN₃ → 2 Na + 3 N₂

You know the following about N₂:

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  • V= 26.5 L
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  • T= 295 K

Replacing in the equation for ideal gas:

1.10 atm* 26.5 L= n* 0.082057 \frac{atm*L}{mol*K}*295 K

Solving:

n=\frac{1.10 atm*26.5 L}{0.082057 \frac{atm*L}{mol*K} *295K}

n= 1.2 moles

Now, the following rule of three can be applied: if 3 moles of N₂ are produced by stoichiometry of the reaction from 2 moles of NaN₃, 1.2 moles of N₂ are produced from how many moles of NaN₃?

moles of NaN_{3}=\frac{1.2 molesofN_{2} *2 molesofNaN_{3} }{3 molesofN_{2} }

moles of NaN₃= 0.8

Since the molar mass of sodium azide is 65.01 g / mol, then one last rule of three applies: if 1 mol has 65.01 grams of NaN₃, 0.8 mol how much mass does it have?

mass of NaN_{3} =\frac{0.8 mol*65.01 grams}{1 mol}

mass of NaN₃=52.008 grams

<u><em>52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.</em></u>

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