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frozen [14]
3 years ago
11

how many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 2.75 g of fe2s3 if the percent yield

for the reaction is 65.0%?
Chemistry
1 answer:
Snezhnost [94]3 years ago
6 0

Answer:

We need 203 mL of FeCl3

Explanation:

Step 1: Data given

molarity FeCl3 = 0.200 M

Mass of Fe2S3 = 2.75 grams

Percent yield = 65.0 %

Step 2: The balanced equation

3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)

Step 3: Calculate moles Fe2S3

Moles Fe2S3 = mass Fe2S3 / molar mass Fe2S3

Moles Fe2S3 = 2.75 grams / 207.9 g/mol

Moles Fe2S3 = 0.0132 moles

Step 4: Calculate theoretical yield

65.0 % = 0.65 = actual yield/ theoretical yield

theoretical yield = actual yield / 0.65

theoretical yield = 0.0132 moles /0.65

theoretical yield = 0.0203 moles Fe2S3

Step 5: Calculate moles FeCl3

For 1 mol Fe2S3 and 6 mol NaCl we need 3 moles Na2S and 2 moles FeCl3

For 0.0203 moles Fe2S3 we need 2*0.0203 = 0.0406 moles FeCl3

Step 6: Calculate volume

Molarity = moles / volume

Volume = moles / molarity

Volume = 0.0406 moles / 0.200 M

Volume = 0.203 L = 203 mL

We need 203 mL of FeCl3

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ELEN [110]
<h2>Answer:HCl</h2>

Explanation:

\text{number of moles}=\frac{\text{given mass}}{\text{molar mass}}

For NH_{3},

\text{given mass}=3g

\text{molar mass}=14+3=17g

n_{NH_{3}}=\frac{3}{17}=0.176

For HCl,

\text{given mass}=5g

\text{molar mass}=1+35.5=36.5g

n_{HCl}=\frac{5}{36.5}=0.136

The reaction between NH_{3} and HCl is

NH_{3}+HCl→NH_{4}Cl

So,one mole of NH_{3} requires one mole of HCl

0.176 moles of NH_{3} requires 0.176 moles of HCl

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HCl will be consumed first.

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7 0
3 years ago
How many chloride ions are in a 220 grams of calcium chloride?
steposvetlana [31]

Answer: Hello, There! Your Answer is Below

2.4 x 1024 ions

Explanation:

220 g of CaCl₂  =  X moles

Solving for X,

                                                X  =  (220 g × 1 mol) ÷ 110.98 g

                                                X  =  1.98 moles

As,

                          1 mole contained  =  1.20 × 10²⁴ Cl⁻ Ions

Then,

                   1.98 mole will contain  =  X Cl⁻ Ions

Solving for X,

                                                 X  =  (1.98 mol × 1.20 × 10²⁴ Ions) ÷ 1mol

                                                 X  =  2.38 × 10²⁴ Cl⁻ Ions

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2 years ago
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Answer:

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So we now have;

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ivann1987 [24]

Answer:

C.

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