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sergeinik [125]
4 years ago
5

Identify the limiting and excess reactants when 1.00 g of zinc reacts with 150 mL of 0.250M Pb(NO3)2. How many grams of lead are

formed in this single replacement reaction
Chemistry
1 answer:
Marysya12 [62]4 years ago
5 0
The reaction equation is:
Zn + Pb(NO₃)₂ → Pb + Zn(NO₃)₂

The moles of lead are calculated using:
moles = concentration x volume
moles = 0.25 x 0.15
moles = 0.0375

The moles of zinc are: 1 / 65 = 0.015

We see from the equation that equimolar quantities of zinc and lead should be present. Therefore, lead is in excess and zinc is the limiting reactant.
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timofeeve [1]

Answer:

0.550

Explanation:

The absorbance (A) of a substance depends on its concentration (c) according to Beer-Lambert law.

A = ε . <em>l</em> . c

where,

ε: absorptivity of the species

<em>l</em>: optical path length

A 45 mM phosphate solution (solution A) had an absorbance of 1.012.

A = ε . <em>l</em> . c

1.012 = ε . <em>l</em> . 45 mM

ε . <em>l</em>  = 0.022 mM⁻¹

We can find the concentration of the second solution using the dilution rule.

C₁ . V₁ = C₂ . V₂

45mM . 11mL = C₂ . 20.0 mL

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The absorbance of the second solution is:

A = (ε . <em>l</em> ). c

A = (0.022 mM⁻¹) . 25 mM = 0.55 (rounding off to 3 significant figures = 0.550)

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