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dem82 [27]
3 years ago
14

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

late the volume of the 1.475 M stock NaOH solution needed to prepare 250.0 mL of 0.1374 M dilute NaOH solution.
Chemistry
1 answer:
Svet_ta [14]3 years ago
8 0

Answer:

V_1=23.3~mL

Explanation:

In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:

C_1*V_1=C_2*V_2

Now we can identify the variables:

C_1=~1.475_M

V_1=~?

C_2=~0.1374_M

V_2=~250.0~mL

If we plug all the values into the equation:

1.475_M*V_1=0.1374_M*250.0~mL

And we solve for V_1:

V_1=\frac{0.1374_M*250.0~mL}{1.475_M}

V_1=23.3~mL

I hope it helps!

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sattari [20]

1) Balance the chemical equation.

3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2

2) List the known and unknown quantities.

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Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4

<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.

<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2

<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2

<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.

Option D.

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1 year ago
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Answer:

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2 years ago
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il63 [147K]

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  <u><em>calculation</em></u>

 Step 1: find moles of ammonia

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8 0
3 years ago
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3 years ago
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BartSMP [9]

Answer:

see explaination

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Check attachment

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