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anyanavicka [17]
2 years ago
10

Ammonium perchlorate is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitrogen gas , ch

lorine gas , oxygen gas , water , and a great deal of energy.
What mass of water is produced by the reaction of of ammonium perchlorate?
Chemistry
1 answer:
Zinaida [17]2 years ago
3 0

Answer:

2.9 g

Explanation:

There is some info missing. I think this is the original question.

<em>Ammonium perchlorate is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitrogen gas, chlorine gas, oxygen gas, water, and a great deal of energy. </em>

<em>What mass of water is produced by the reaction of </em><em>9.6 g</em><em> of ammonium perchlorate?</em>

<em />

Step 1: Given data

Mass of ammonium perchlorate: 9.6 g

Step 2: Write the balanced equation

2 NH₄ClO₄(s) → N₂(g) + Cl₂(g) + 2 O₂(g) + 4 H₂O(l)

Step 3: Calculate the moles corresponding to 9.6 g of ammonium perchlorate

The molar mass of ammonium perchlorate is 117.49 g/mol.

9.6 g \times \frac{1mol}{117.49 g} = 0.082mol

Step 4: Calculate the moles of water formed from 0.082 moles of ammonium perchlorate

The molar ratio of NH₄ClO₄ to H₂O is 2:4. The moles of water formed are 4/2 × 0.082 mol = 0.16 mol

Step 5: Calculate the mass corresponding to 0.16 moles of water

The molar mass of water is 18.02 g/mol.

0.16 mol \times \frac{18.02g}{mol} = 2.9 g

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a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh
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Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

Molarity = M = \frac{1,22 mol solute}{lts solution}

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

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1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%

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702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent  

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Finally, we have:

Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03

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