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vagabundo [1.1K]
3 years ago
5

Sodium reacts with chlorine gas to form sodium chloride. Write a balanced chemical equation for the reaction, and find the mass

of chlorine gas that will react with 96.6 g of sodium.
Chemistry
2 answers:
weqwewe [10]3 years ago
8 0
2Na + Cl2 ⇒ 2NaCl

96.6g Na ((1 mole Na)/(22.99 g Na))((1 mole Cl2)/(2 mole Na))((70.9 g Cl2)/(1 mole Cl2))=149 g Cl2

Bess [88]3 years ago
4 0
Chemical equation:
2 Na + Cl _{2}   -\ \textgreater \  2NaCl
M ( Na ) = 23 g/moles   ( 2 moles Na = 46 g )
M ( Cl 2 ) = 71 g/moles  ( 1 mole Cl 2 = 71 g )
so Na needs more Cl 2 to react;
96  : x = 46 : 71
46 x = 6816
x = 6816 : 46
x = 148.17 g
Answer: the mass of chlorine gas is 148.17 g.
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If water was used to rinse the burette used to dispense the hydrochloric acid solution, how would this affect
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Answer:

The concentration of hydrochloric acid would be estimated to be less.

Explanation:

This is because, the hydroxyl ions from water react with the hydrogen ions from the hydrochloric acid, hence decreasing the moles of hydrogen ions which lowers the acidic strength of Hydrochloric acid.

7 0
2 years ago
A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
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Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

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6 0
3 years ago
The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.376 g sample of e
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Answer:

The answer to your question is: C₄H₁₀O

Explanation:

Data

          CxHyOz

mass sample : 1.376 g

mass CO₂ = 3.268 g

mass H₂O = 1.672 g

Process

Reaction

                      CxHyOz  + O₂ ⇒   CO₂  +  H₂O

1.- Calculate the moles and mass of carbon

Molecular mass CO₂ = 44g

                      44 g of CO₂ --------------  12 g of C

                      3.268 g of CO₂  --------    x

                         x = (3.268 x 12) / 44

                        x = 0.891 g of Carbon

                       12 g of carbon -----------  1 mol

                       0.891 g of C     ----------   x

                       x = (0.891 x 1) / 12

                       x = 0.0743 moles of carbon

2.- Calculate the moles and mass of hydrogen

                      18 g of water --------------- 2 g of H

                      1.672 g of H₂O ------------  x

                      x = (1.672 x 2) / 18

                      x = 0.186 g of hydrogen

                      1 g of hydrogen ------------  1 mol of H

                      0.186 g of H       ------------  x

                      x = (0.186 x 1) / 1

                      x = 0.186 moles of H

3.- Calculate the mass of Oxygen and its moles

Mass of Oxygen = 1.376 - 0.891 - 0.186

                           = 0.299 g of O₂

Moles of Oxygen

                             16 g of Oxygen ---------------- 1 mol

                             0.299 g of O    -----------------  x

                             x = (0.299 x 1) / 16

                             x = 0.019 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon         0.0743/ 0.019 = 3.9 ≈ 4.0

Hydrogen     0.186/ 0.019 = 9.7 = 10

Oxygen         0.019/ 0.019 = 1

5.- Write the empirical formula

                              C₄H₁₀O                  

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