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IgorLugansk [536]
3 years ago
10

Một hỗn hợp X gồm CH3OH; C2H5OH; phenol có khối lượng 28,9 gam phản ứng vừa hết với 100ml dung dịch NaOH 2M. Phần trăm theo khối

lượng phenol là ? Biết C = 12; O = 16; H = 1; Na = 23. *
Chemistry
1 answer:
slavikrds [6]3 years ago
3 0

Answer:

nNaOH = 0,2 mol

Trong hỗn hợp các chất đề bài cho chỉ có phenol tác dụng với NaOH

C6H5OH + NaOH → C6H5ONa + H2O

   0,2  ←     0,2 (mol)

⟹ mphenol = 0,2.94 = 18,8 gam

⟹  

%mC6H5OH=18,8/28,9 x100%= 65,05%

Explanation:

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phosphorous- 5

calcium- 2

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3 0
2 years ago
Suppose that 0.410 mol of methane, CH4(g), is reacted with 0.560 mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole produc
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Answer:

The balanced equation for this reaction will be

                            CH4 + 4F2    →  CF4 + 4HF

We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2

So using the unitary method we will get that

  • 1 mole of CH4 → 4 mole of 4 mole of fluorine
  • 0.41 mole of methane  →  4*0.41 = 1.64 mole of fluorine for complete reaction

but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.

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  • 0.56 mole →  \frac{1}{4} * 0.56 = 0.14mole of CF4
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now to find the heat released we have the formula as

DELTA H = n * Delta H of product - n *delta H of reactant

where n is the moles of the reactant and product.

note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.

8 0
3 years ago
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