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Assuming an excess of all other necessary reagents, how many grams of acetone semicarbazone can be prepared from 0.25 grams of acetone? Enter only the number with two significant figures. (acetone semicarbazone 115.14 g/mol, acetone 58.08 g/mol)

Respuesta :

Answer:

We can prepare 0.50 grams of semi carbazone

Explanation:

Step 1: Data given

Acetone semicarbazone = C4H9N3O

Acetone = C3H6O

Molar mass C4H9N3O = 115.14 g/mol

Mass of acetone = 0.25 grams

Molar mass acetone = 58.08 g/mol

Step 2: The balanced equation

CH6ClN3O + C3H6O + C2H3NaO2 → C4H9N3O + NaCl + H2O + C2H4O2

Step 3: calculate moles of acetone

Moles acetone = mass acetone / molar mass acetone

Moles acetone = 0.25 grams / 58.08 g/mol

Moles acetone = 0.0043 moles

Step 4: Calculate moles acetone semicarbazone

Mole ratio is 1:1

For 0.0043 moles acetone we'll have 0.0043 moles acetone semicarbazone

Step 5: Calculate mass semi carbazone

Mass C4H9N3O = moles C4H9N3O * molar mass C4H9N3O

Mass C4H9N3O  = 0.0043 moles * 115.14 g/mol

Mass C4H9N3O = 0.50 grams

We can prepare 0.50 grams of semi carbazone

Eduard22sly

Answer: 0.50g of acetone semicarbazone will be produced.

Explanation:

CH3COCH3 + NH2NHCONH2 —> (CH3)2CNNHCONH2

Molar Mass of acetone semicarbazone = 115.14g

Molar Mass of acetone = 58.08g

From the equation,

58.08g of acetone produced 115.14g of acetone semicarbazone.

Therefore, 0.25 grams of acetone will produce = (0.25x115.14) /58.08 = 0.50g of acetone semicarbazone

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