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Novay_Z [31]
3 years ago
11

What is the concentration of nitrate ion in a 425 mL solution containing 32.0 g of Mg(NO3)2 (M = 148.3)?

Chemistry
1 answer:
Butoxors [25]3 years ago
6 0
Mg(NO3)2 ➡️ Mg2+ + 2 NO3-

(32.0g Mg(NO3)2) / (148.3g Mg(NO3)2/mol)* (2 mol NO3- / 1 mol Mg(NO3)2) / (0.425 L) = 1.02 mol/ L NO3-
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If the rate law for a reaction A → P is rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122 M, calcula
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An alloy is a mixture that has
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Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M
Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

  • 2AgNO₃(aq) + K₂CrO₄(aq)  → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to <em>identify the limiting reactant</em>:

We have:

  • 0.20 M * 50.0 mL = 10 mmol of AgNO₃
  • 0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. <em><u>That makes K₂CrO₄ the limiting reactant</u></em>.

Now we <u>calculate the mass of Ag₂CrO₄ formed</u>, using the <em>limiting reactant</em>:

  • 4 mmol K₂CrO₄ * \frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4} = 1326.92 mg Ag₂CrO₄
  • 1326.92 mg / 1000 = 1.327 g Ag₂CrO₄
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3 years ago
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