The minimum volume of 0.2M sodium sulfide that will precipitate out aluminum from 50.0 mL, 0.25 M aluminum nitrate would be 0.094 L or 94 mL
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
Mole ratio of Na2S and Al(NO3)3 = 3:2
Mole of 50.0 mL, 0.25 M Al(NO3)3 = 50/1000 x 0.25
= 0.0125 mole
Equivalent mole of Na2S = 3/2 x 0.0125
= 0.0188 mole
Volume of 0.2M, 0.0188 mole Na2S = 0.0188/0.2
= 0.094 L or 94 mL
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Answer:
1.4 × 10^-4 M
Explanation:
The balanced redox reaction equation is shown below;
5Fe2+ + MnO4- + 8H+ --> 5Fe3+ +Mn2+ + 4H2O
Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol
Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles
Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M
Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M
Volume of Fe^2+ (VA)= 20.0 ml
Let the concentration of MnO4^- be CB (the unknown)
Volume of the MnO4^- (VB) = 12.6 ml
Let the number of moles of Fe^2+ be NA= 5 moles
Let the number of moles of MnO4^- be NB = 1 mole
From;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CB= CAVANB/VBNA
CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5
CB = 1.4 × 10^-4 M
acid
when a metal hydroxide reacts with acid it forms a salt and water, for example:
lithium hydroxide + hydrochloric acid → lithium chloride (the salt) + water
