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2 years ago

NEED HELP ASAP NOT DIFFICULT

Chemistry

2 answers:

Jet001 [13]2 years ago

7 0

Answer:

9 moles of NaNO3 and 3 moles of AlCl3

Explanation:

First, we need the balanced equation, which is:

Al(NO3)3 + (3)NaCl ---> (3)NaNO3 + AlCl3

Next, we need the limiting reactant, which is NaCl because we require more of it than Al(NO3) to make AICl

Now we need to determine the maximum of AlCl we can produce:

Because we need 3 moles of NaCl to react with Al(NO3)3 to produce 1 mole of AlCl3, we can only produce 3 AlCl3 due to insufficient NaCl.

So we would have 9 NaNO3 and 3 AlCl3.

Burka [1]2 years ago

5 0

Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

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Ideal He gas expanded at constant pressure of 3 atm until its volume was increased from 9 liters to 15 liters. During this proce

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Explanation:

The given data is as follows.

P = 3 atm

= $3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}$

= $3.03975 \times 10^{5} Pa$

$V_{1}$ = 9 L = $9 \times 10^{-3} m^{3}$ (as 1 L = 0.001 $m^{3}$ ),

$V_{2}$ = 15 L = $15 \times 10^{-3} m^{3}$

Heat energy = 800 J

As relation between work, pressure and change in volume is as follows.

W = $P \times \Delta V$

or, W = $P \times (V_{2} - V_{1})$

Therefore, putting the given values into the above formula as follows.

W = $P \times (V_{2} - V_{1})$

= $3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})$

= 1823.85 Nm

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3 years ago

Submit your answer for the remaining reagent in Tutorial Assignment #1 Question 9 here, including units. Note: Use e for scienti

djverab [1.8K]

<h3>Answer:</h3>

The mass of excessive water (H₂O) is 40.815 kg

<h3>Explanation:</h3>

The Equation for the reaction is;

Li₂O(s) + H₂O(l) → 2LiOH(s)

From the question;

Mass of water removed is 80.0 kg

Mass of available Li₂O is 65.0 kg

We are required to calculate the mass of excessive reagent.

<h3>Step 1: Calculating the number of moles of water to be removed</h3>

Moles = Mass ÷ Molar mass

Molar mass of water = 18.02 g/mol

Mass of water = 80 kg (but 1000 g = 1kg)

= 80,000 g

Therefore;

$Moles of water = \frac{80,000g}{18.02 g/mol}$

= 4.44 × 10³ moles

<h3>Step 2: Moles of Li₂O available </h3>

Moles = mass ÷ molar mass

Mass of Li₂O available = 65.0 kg or 65,000 g

Molar mass Li₂O = 29.88 g/mol

Moles of Li₂O = 65,000 g ÷ 29.88 g/mol

= 2.175 × 10³ moles Li₂O

<h3>Step 3: Mass of excess reagent </h3>

From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)

1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH

The ratio of Li₂O to H₂O is 1:1

Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
However, the number of moles of water to be removed is 4.44 × 10³ moles but only 2.175 × 10³ moles will react with the available Li₂O.
This means, Li₂O is the limiting reactant while water is the excessive reagent.

Therefore:

Moles of excessive water = 4.44 × 10³ moles - 2.175 × 10³ moles

= 2.265 × 10³ moles

Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol

= 4.0815 × 10⁴ g or

= 40.815 kg

Thus, the mass of excessive water is 40.815 kg

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3 years ago