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Calculate the number of milliliters of 0.538 M NaOH required to precipitate all of the Fe2+ ions in 198 mL of 0.552 M FeBr2 solution as Fe(OH)2. The equation for the reaction is:FeBr2(aq) + 2NaOH(aq) Fe(OH)2(s) + 2NaBr(aq)

Question

Calculate the number of milliliters of 0.538 M NaOH required to precipitate all of the Fe2+ ions in 198 mL of 0.552 M FeBr2 solution as Fe(OH)2. The equation for the reaction is:

FeBr2(aq) + 2NaOH(aq) Fe(OH)2(s) + 2NaBr(aq)

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Step 1

Given,

198 ml of 0.552 M of FeBr...

Given reaction is shown below
FeBr, (aq) 2NaOH(aq)
Fe(OH)2 (s) + 2NaBr(aq)
Moles of FeBr2 is calculated as follows,
Number of moles= Volumex Molarity
198
-x0.5 52 M
1000
Number of moles of FeBr,-
Number of moles of FeBr,
0.1092
According to the reaction,
one mole of FeB,reaction with two moles sodium hydroxide. Therefore,
number of moles of N2OH requires is given below,
Number of moles of NaOH required=2x0.1092
Number of moles of NaOH required 0.2186
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Given reaction is shown below FeBr, (aq) 2NaOH(aq) Fe(OH)2 (s) + 2NaBr(aq) Moles of FeBr2 is calculated as follows, Number of moles= Volumex Molarity 198 -x0.5 52 M 1000 Number of moles of FeBr,- Number of moles of FeBr, 0.1092 According to the reaction, one mole of FeB,reaction with two moles sodium hydroxide. Therefore, number of moles of N2OH requires is given below, Number of moles of NaOH required=2x0.1092 Number of moles of NaOH required 0.2186

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