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Stolb23 [73]
2 years ago
12

Amino axit X có công thức H2NCxHy(COOH)2. Cho 0,1 mol X vào 0,2 lít dung dịch H2SO4 0,5M, thu được dung dịch Y. Cho Y phản ứng v

ừa đủ với dung dịch gồm NaOH 1M và KOH 3M, thu được dung dịch chứa 36,7 gam muối. Phần trăm khối lượng của nitơ trong X là
Chemistry
1 answer:
Taya2010 [7]2 years ago
7 0

Answer:

nH2SO4 = 0,1 mol

Đặt nNaOH = a; nKOH = 3a (mol)

Quy đổi phản ứng thành: {X, H2SO4} + {NaOH, KOH} → Muối + H2O

Ta có: nH+ = nOH- → 2nX + 2nH2SO4 = nNaOH + nKOH

→ 2.0,1 + 2.0,1 = a + 3a → a = 0,1

→ nH2O = nH+ = nOH- = 0,4 mol

BTKL: mX + mH2SO4 + mNaOH + mKOH = m muối + mH2O

→ mX + 0,1.98 + 0,1.40 + 0,3.56 = 36,7 + 0,4.18 → mX = 13,3 gam

→ MX = 13,3/0,1 = 133

→ %mN = (14/133).100% ≈ 10,526%

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Answer:

Option D. 30 g

Explanation:

The balanced equation for the reaction is given below:

2Na + S —> Na₂S

Next, we shall determine the masses of Na and S that reacted from the balanced equation. This is can be obtained as:

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 g

Molar mass of S = 32 g/mol

Mass of S from the balanced equation = 1 × 32 = 32 g

SUMMARY:

From the balanced equation above,

46 g of Na reacted with 32 g of S.

Finally, we shall determine the mass sulphur, S needed to react with 43 g of sodium, Na. This can be obtained as follow:

From the balanced equation above,

46 g of Na reacted with 32 g of S.

Therefore, 43 g of Na will react with = (43 × 32)/46 = 30 g of S.

Thus, 30 g of S is needed for the reaction.

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