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WITCHER [35]
3 years ago
7

What is the molarity of aqueous lithium bromide if 25.0 mL of LiBr reacts with 10.0 mL of 0.250 M Pb(NO 3) 2

Chemistry
1 answer:
bixtya [17]3 years ago
3 0

The molarity of aqueous lithium bromide, LiBr solution is 0.2 M

We'll begin by calculating the number of mole of Pb(NO₃)₂ in the solution.

  • Volume = 10 mL = 10 / 1000 = 0.01 L
  • Molarity of Pb(NO₃)₂ = 0.250 M
  • Mole of Pb(NO₃)₂ =?

Mole = Molarity x Volume

Mole of Pb(NO₃)₂ = 0.25 × 0.01

Mole of Pb(NO₃)₂ = 0.0025 mole

Next, we shall determine the mole of LiBr required to react with 0.0025 mole of Pb(NO₃)₂

Pb(NO₃)₂ + 2LiBr —> PbBr₂ + 2LiNO₃

From the balanced equation above,

1 mole of Pb(NO₃)₂ reacted with 2 mole of LiBr.

Therefore,

0.0025 mole of Pb(NO₃)₂ will react with = 2 × 0.0025 = 0.005 mole of LiBr

Finally, we shall determine the molarity of the LiBr solution

  • Mole = 0.005 mole
  • Volume = 25 mL = 25 / 1000 = 0.025 L
  • Molarity of LiBr =?

Molarity = mole / Volume

Molarity of LiBr = 0.005 / 0.025

Molarity of LiBr = 0.2 M

Learn more about molarity: brainly.com/question/10103895

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How many moles are in 7.6 g of Ca
klasskru [66]

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The Surface of Earth is changing slowly<br> over time.A.False<br> B.true<br> Previous
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7 0
3 years ago
If 23.6 g of hydrogen gas reacts with 28.3 g of nitrogen gas, what is the maximum amount of product that can be produced?
Aleonysh [2.5K]

Answer:

34.3 g NH3

Explanation:

M(H2) = 2*1 = 2 g/mol

M(N2) = 2*14 = 28 g/mol

M(NH3) = 14 + 3*1 = 17 g/mol

23.6 g H2* 1 mol/2 g = 11.8 mol H2

28.3 g N2 * 1 mol/28 g = 1.01 mol N2

                                 3H2 + N2 ------> 2NH3

from reaction         3 mol    1 mol

given                   11.8 mol    1.01 mol

We can see that H2 is given in excess, N2 is limiting reactant.

                                 3H2 + N2 ------> 2NH3

from reaction                     1 mol         2 mol

given                                 1.01 mol      x

x = 2*1.01/1= 2.02 mol NH3

2.02 mol * 17g/1 mol ≈ 34.3 g NH3

8 0
3 years ago
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