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Alecsey [184]
3 years ago
14

Combining 0.380 mol Fe2O3 with excess carbon produced 13.7 g Fe

Chemistry
1 answer:
Alex_Xolod [135]3 years ago
8 0

Answer:

Percent yield = 32.9%

Explanation:

Given data:

Number of moles of Fe₂O₃ = 0.380 mol

Number of moles of Fe produced = 13.7 g

Theoretical yield = ?

Actual yield of iron in mol = ?

Percent yield = ?

Solution:

Chemical equation:

Fe₂O₃ + 3C   →   2Fe  + 3CO

Actual yield of iron in moles:

Number of  moles = mass/ molar mass

Number of moles = 13.7 g/ 55.8 g/mol

Number of moles = 0.25 mol

Theoretical yield:

                    Fe₂O₃          :           Fe

                      1                :              2

                    0.380         :          2×0.380 = 0.76

Theoretical yield = 0.76 mol

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 0.25 mol /0.76 mol × 100

Percent yield = 32.9%

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Which of the following solids would not decompose on heating​
Gekata [30.6K]

Answer:

Anhydrous sodium carbonate is stable to heat and does not decompose even when it is heated to redness. This is because sodium carbonate salt on heating with acids react to release carbon dioxide.

5 0
3 years ago
10. Isolation of a pure sample of the third product, which has been determined to be an isomer of the major and minor products,
Ymorist [56]

Answer:

Four possible isomers (1–4) for the natural product essramycin. The structure of compound 1 was attributed to essramycin by 1H NMR, 13C NMR, HMBC, HRMS, and IR experiments.

Explanation:

Three synthetic routes were used to prepare all four compounds (Figure 2A). All three reactions utilize 2-(5-amino-4H-1,2,4-triazol-3-yl)-1-phenylethanone (5) as the precursor, whereas each uses different esters (6–8) to construct the pyrimidinone ring. Isomer 1 was prepared by reaction A, which used triazole 5 and ethyl acetoacetate (6) in acetic acid. This was the reaction used in syntheses of essramycin by the Cooper and Moody laboratories.3,4 Reaction B produced compound 2 (minor product) and compound 3 (major product), which were separated chromatographically. This reaction allowed reagent 5 to react with ethyl 3-ethoxy-2-butenoate (7) in the presence of sodium in methanol, under reflux for 24 h. Compound 4 was prepared by reaction C, which was obtained by reflux of 5 and methyl 2-butynoate (8) in n-butanol.

7 0
3 years ago
How can scientists detect blood doping?
aev [14]

Answer:

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6 0
2 years ago
Read 2 more answers
Consider a situation in which 211 g
Stella [2.4K]

Answer:

3.00 mol

Explanation:

Given data:

Mass of P₄ = 211 g

Mass of oxygen = 240 g

Moles of P₂O₅ = ?

Solution:

Chemical equation:

P₄ + 5O₂       →     2P₂O₅

Number of moles of P₄:

Number of moles = mass/ molar mass

Number of moles = 211 g / 123.88 g/mol

Number of moles = 1.7 mol

Number of moles of O₂ :

Number of moles = mass/ molar mass

Number of moles = 240 g / 32g/mol

Number of moles = 7.5 mol

Now we will compare the moles of product with reactant.

                       O₂         :         P₂O₅

                        5          :           2

                        7.5       :        2/5×7.5 = 3.00

                       P₄          :         P₂O₅

                        1           :           2

                       1.7         :       2×1.7 = 3.4 mol

Oxygen is limiting reactant so the number of moles of P₂O₅ are 3.00 mol.

Mass of P₂O₅:

Mass = number of moles × molar mass

Mass = 3 mol ×283.9 g/mol

Mass = 852 g

3 0
3 years ago
The pressure of a 70.0L sample of gas is 600 mm Hg at 20.0C. If the temperature drops to 15.0C and the volume expands to 90.0L,
Mekhanik [1.2K]

Answer:

458.7 mmHg

Explanation:

Step 1:

Data obtained from the question. This includes:

Initial volume (V1) = 70L

Initial pressure (P1) = 600 mmHg

Initial temperature (T1) = 20°C

Final temperature (T2) = 15°C

Final volume (V2) = 90L

Final pressure (P2) =...?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This is illustrated below:

T(K) = T (°C) + 273

Initial temperature (T1) = 20°C

Initial temperature (T1) = 20°C + 273 = 293K

Final temperature (T2) = 15°C

Final temperature (T2) = 15°C + 273 = 288K

Step 3:

Determination of the new pressure of the gas.

The new pressure of the gas can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

600 x 70/293 = P2 x 90/288

Cross multiply to express in linear form

P2 x 90 x 293 = 600 x 70 x 288

Divide both side by 90 x 293

P2 = (600 x 70 x 288) / (90 x 293)

P2 = 458.7 mmHg

Therefore, the new pressure of the gas is 458.7 mmHg

5 0
3 years ago
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