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tankabanditka [31]
3 years ago
15

A 0.964 gram sample of a mixture of sodium formate and sodium chloride is analyzed by adding sulfuric acid. The equation for the

reaction for sodium
Chemistry
1 answer:
frosja888 [35]3 years ago
5 0

Answer: 67.8 %.

Explanation:

Okay, let us delve right into the solution to the question;

The balanced chemical reaction is given by the equation (1) below;

2 HCOONa + H2SO4 ---------> 2 CO + 2 H2O + Na2SO4. ----------------------------------------------------------------------------(1).

From the balanced chemical reaction in equation (1) above we can see that; 2 moles of HCOONa reacts with one moles of tetraoxosulphate acid, H2SO4 to produce 2 moles of carbonmonoxide,CO; 2 moles of water, H2O and 1 mole of sodium tetraoxosulphate, Na2SO4.

The parameters given from the question are; total atmospheric pressure, P(t) = 752 torr, volume of CO= 242 mL = 0.242 Litres.

STEP ONE : find the carbon monoxide,CO pressure; P(CO).

Using the formula below;

P(t) = P(CO) + P(H2O). Hence;

P(CO) = P(t) - P(H2O). Note that P(H2O)= 19.8 torr.

==>P(CO)= 752 torr - 19.8 torr = 732.2 torr.

STEP TWO: calculate the number of moles of Carbonmonoxide,CO.

Using the formula below;

Number of moles= pressure(P) × volume(v) / gas constant(R) × temperature (T).

That is, n= PV/RT.

n= 732 torr × 0.242 Litres/ 62.4 × 295.15.

= 9.62 × 10^-3 mol of CO.

STEP THREE:

2 moles of HCOONa = 2 moles of CO.

=> 2 moles of HCOONa = 2 moles of CO/ 2 moles of CO = 1 mol( HCOONa/ CO).

Then, 9.62 × 10^-3 mol of CO × 1 mol( HCOONa/ CO).

==> 9.62 × 10^-3 mol HCOONa × molar masss of HCOONa(68 grams/mol)

= 0.654 grams.

Therefore, the percentage of sodium formate in the original mixture = 0.654 grams/ 0.964 gram × 100 = 67.8 %.

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What is the balanced equation for the reaction of a solution of sodium sulfate is mixed with strontium chloride?
natta225 [31]

Answer:

NaSO⁴(ads) ,ganadicate

6 0
3 years ago
A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this
NeTakaya

Answer:

a)  molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane CH_4}= 16 g/mol

Molar mass of ethane C_2H_6= 30 g/mol

Molar mass of ethylene C_2H_4 = 28 g/mol

Molar mass of water H_2O=18g/mol

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

Their molar composition can be calculated as follows:

n_{CH_4}= \frac{75}{16}

n_{CH_4}= 4.69 moles

n_{C_2H_6} = \frac{10}{30}

n_{C_2H_6} = 0.33 moles

n_{C_2H_4} = \frac{5}{28}

n_{C_2H_4} = 0.18 moles

n_{H_2O}= \frac{10}{18}

n_{H_2O}= 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = \frac{n_{H_2O}}{n_{drygas}}

= \frac{0.56}{5.2}

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

    CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O

4.69         2× 4.69

moles       moles

   C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O

0.33      3.5 × 0.33

moles    moles

    C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O

0.18           3× 0.18

moles        moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h

Total moles of O_2 required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles O_2

Thus, moles of air required = \frac{1}{0.21}*11.075

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

5 0
3 years ago
FORENSIC SCIENCE HELP
Annette [7]
That is correct c
Explanation
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Which option explains how some parts of a plant can be white?
just olya [345]

Some parts of a plant can be white because they do not contain chlorophyll.

Chlorophyll is a term to refer to a characteristic compound of most plants that has the function of:

  • Bring the green color to vegetables
  • Take charge of the photosynthesis process.

Some plants do not have as much chlorophyll in their systems. However, this does not affect their reproduction and growth because they have developed other feeding methods through their roots.

These plants that contain low amounts of chlorophyll have pale and white colorations because chlorophyll is what gives the plants their intense green color.

Note: This question is incomplete because options are missing. Nevertheless, I can answer it based on my prior knowledge.

Learn more in: brainly.com/question/14884020

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2 years ago
What was the main reason why Alfred Wegener's theory about the movement of the continents was dismissed in 1912?
eduard

Answer:

His map of the ocean floor did not support his theory

Explanation:

  • Alfred Wegner was a great scientist
  • He observed the plate tectonics of earth and described it in a theory.
  • It was dismissed after viewing his ocean floor
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2 years ago
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