Question
Asked Oct 16, 2019
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How many moles of precipitate will be formed when 108.5 mL of 0.200 M NaBr is reacted with excess Pb(NO₃)₂ in the following chemical reaction? 2 NaBr (aq) + Pb(NO₃)₂ (aq) → PbBr₂ (s) + 2 NaNO₃ (aq)
 
 
 
 
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Expert Answer

Step 1

Given,

Volume of NaBr = 108.5 mL = 0.1085 L       (1 mL =0.001 L)

Molarity of NaBr = 0.200 M = 0.2 mol/L

Moles of NaBr can be calculated as:

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Moles MolarityVolume in litres Moles Molarity x Volume in litres Moles 0.2 mol/L x 0.1085 L Moles 0.0217 mol

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Step 2

The balance reaction given is:

2 NaBr (aq) + Pb(NO3)2 (aq) → PbBr2 (s) + 2 NaNO3 (aq)

Here, PbBr2 (s) is the ...

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1 mol PbBr2 2 mol NaBr reacts to give 1 mol PbBr2 x 0.0217 mol NaBr : 0.0217 mol NaBr reacts to give 2 mol NaBr - 0.01085 mol PbBr2

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