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KatRina [158]
3 years ago
15

how many moles of aluminum are needed to produce 0.418 mol of Al2(SO4)3? 2 Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)

Chemistry
1 answer:
nlexa [21]3 years ago
5 0
<h3>Answer:</h3>

0.836 mol Al

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Al (s) + 3H₂SO₄ (aq) → Al₂(SO₄)₃ (aq) + 3H₂ (g)

[Given] 0.418 mol Al₂(SO₄)₃

[Solve] <em>x</em> mol Al

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Al (s) → 1 mol Al₂(SO₄)₃ (aq)

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                      \displaystyle 0.418 \ mol \ Al_2(SO_4)_3(\frac{2 \ mol \ Al}{1 \ mol \ Al_2(SO_4)_3})
  2. [DA] Multiply/Divide [Cancel out units]:                                                            \displaystyle 0.836 \ mol \ Al

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

Since our final answer already has 3 sig figs, there is no need to round.

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nasty-shy [4]

Answer:

0.1440M

Explanation:

Let''s bring out the parameters we were given;

Rate constant = 8.74 x 10^-4s^-1

Initial Concentration [A]o = 0.330M

Final concentration [A]= ?

Time = 800s

The reaction is a first order reaction, due to the unit of the rate constant. In first order reactions, the reaction rate is directly proportional to the reactant concentration and the units of first order rate constants are 1/sec.

Formular relating these parameters is given as;

ln[A] = ln[A]o − kt

Making [A] subject of interest, we have;

ln[A] = ln[A]o − kt

ln[A]  = ln(0.330) - ( 8.74 x 10^-4 * 800)

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ln[A] = -1.8079

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8 0
3 years ago
A water treatment plant has 4 settling tanks that operate in parallel (the flow gets split into 4 equal flow streams), and each
ale4655 [162]

Answer:

a) When the 4 tanks operate in parallel the retention time is 1.26 hours.  

b) If the tanks are in series, the retention time would be 0.31 hours

Explanation:

The plant has 4 tanks, each tank has a volume V = 600 m_3. The total flow to the plant is Ft = 12 MGD (Millions of gallons per day)

When we use the tanks in parallel, it means that the total flow will be divided in the total number of tanks. F1 will be the flow of each tank.

Firstly, we should convert the MGD to m_3 /day. In that sense, we can calculate the retention time using the tank volume in m_3.

Ft =12 \frac{MG}{day}  (\frac{1*10^6 gall}{1MG} ) (\frac{3.78 L}{1 gall} ) (\frac{1 dm^3}{1L} ) (\frac{1 m^3}{10^3 dm^3} ) = 45360 \frac{m^3}{day}

After that, we should divide the total flow by four, because we have four tanks.

F1 = Ft/4 =(45360 \frac{m^3}{d} )/4 = 11 340 \frac{m^3}{d}

To calculate the retention time we divide the total volume V by the flow of each tank F1.

t1 =\frac{V1}{F1} = \frac{600 m^3}{11340  \frac{m^3}{d} }  = 0.0529 day\\t1 = 0.0529 d (\frac{24 h}{1d} ) = 1.26 h

After converting t1 to hours we found that the retention time when the four reactors are in parallel is 1.26 hours.

b)

If the four reactors were working in series, the entire flow goes first through one tank, then the second and so on. It means the total flow will be the flow of each tank.

In that order of ideas, the flow for reactors in series will be F2, and will have the same value of F0.

F2 = F0

To calculate the retention time t2 we divide the total volume V by the flow of each tank F2.

t2 =\frac{V1}{F2} = \frac{600 m^3}{45360  \frac{m^3}{d} }  = 0.01 day\\t2 = 0.01  d (\frac{24 h}{1d} ) = 0.31 h

After converting t2 to hours we found that the retention time when the four reactors are in series is 0.31 hours.

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