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Over [174]
3 years ago
11

suppose that 0.05 g of NaHCO3 reacts with 0.05 g of HCI and the reaction bubbles due to the carbon dioxide being produced. if th

e substance produced are 0.04 g water and 0.03g of salt how many grams if carbon dioxide gas were produced
Chemistry
1 answer:
VLD [36.1K]3 years ago
6 0

The balanced chemical equation for the given reaction:

NaHCO₃ + HCl --> NaCl + H₂O + CO₂

Given, mass of NaHCO₃ = 0.05 g

Molar mass of NaHCO₃ = 84.007 g/mol

Moles = mass/molar mass

Moles of NaHCO₃ = 0.05 g/84.007 g/mol

Moles of NaHCO₃ = 6.0 x 10⁻⁴

Given, mass of HCl = 0.05 g

Molar mass of HCl = 36.46 g/mol

Moles of HCl = 0.05 g/ 36.46 g/mol

Moles of HCl = 1.4 x 10⁻³

Since NaHCO3 has the least number of moles, it is the limiting reagent.

As per the balanced equation the molar ratio between NaHCO₃ : CO₂ is 1:1

Therefore, moles of CO₂ = 6.0 x 10⁻⁴

Molar mass of CO₂ = 44 g/mol

Mass = moles x molar mass

Mass of CO₂ =  6.0 x 10⁻⁴ mol x 44 g/mol = 0.0264 g

Mass of carbon dioxide gas produced = 0.0264 g

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Please help with chemistry question... It would help me tremendously and if you can, please show work :)?
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Answer:

5.06 atm

Explanation:

Step 1:

Data obtained from the question. This includes:

Mass of S2O = 175g

Volume (V) = 16600 mL

Temperature (T) = 195°C

Pressure (P)

Step 2:

Determination of the number of mole of S2O in 175g of S2O.

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Number of mole of S2O =.?

Number of mole = Mass/Molar Mass

Number of mole of S2O = 175/80

Number of mole of S2O = 2.1875 moles

Step 3:

Conversion to appropriate units.

It is essential to always express the various variables in the right units of measurement in order to obtain the desired answer in the right units.

For volume:

1000mL = 1L

Therefore, 16600mL = 16600/1000 = 16.6L

For temperature:

Temperature (Kelvin) = temperature (celsius) + 273

Temperature (celsius) = 195°C

Temperature (Kelvin) = 195°C + 273 = 468K

Step 4:

Determination of the pressure.

The pressure can be obtained by the application of the ideal gas equation. This is illustrated below:

Volume (V) = 16.6L

Temperature (T) = 468K

Number of mole (n) = 2.1875 moles

Gas constant (R) = 0.082atm.L/Kmol

Pressure (P) =

PV = nRT

P x 16.6 = 2.1875 x 0.082 x 468

Divide both side by 16.6

P = (2.1875 x 0.082 x 468) /16.6

P = 5.06 atm

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student in chemistry 150-02 weighed out 55.5g of octane c8h18 and allowed it to react with oxygen, o2 the product formed were ca
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Answer:

C8H8 + 10O2 → 8CO2 + 4H2O

Explanation:

unbalanced reaction:

C8H8 + O2 → CO2 + H2O

balanced for semireactions:

(1) 16H2O + C8H8 → 8CO2 + 40H+

(2) 10(4H+ + O2 → 2H2O)

⇒ 40H+ + 10O2 → 20H2O

(1) + (2):

balanced reaction:

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                     20 - O2 - 20

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