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Biology and Chemistry Homework Help
How do you calculate moles to neutralise oxalic acid?
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[QUOTE="Daniel2244, post: 6134007, member: 640488"] [h2]Homework Statement [/h2] calculate the number of moles of potassium manganate required to neutralise oxalic acid. 2KMn)[SUB]4[/SUB][SUP]-[/SUP] + 5H[SUB]2[/SUB]C[SUB]2[/SUB]O[SUB]4[/SUB][SUP]2-[/SUP]+ 16H[SUP]+[/SUP] --> 2Mn[SUP]2+[/SUP] + 10CO[SUB]2[/SUB] + H[SUB]2[/SUB]O [h2]Homework Equations[/h2] n = c x V n= m/Ar [h2]The Attempt at a Solution[/h2] Moles of oxalic acid = 5x10[SUP]-4[/SUP] Volumeol of KMnO[SUB]4[/SUB] = 0.01085dm[SUP]-3[/SUP] Concentration of oxalic acid = 0.05 therefore concentration of manganate is the same (might be wrong). moles manganate= c x V = 0.05 X 5x10[SUP]-4[/SUP]= 5.425x10[SUP]-4[/SUP]. However, I think this is wrong and that you're meant to use the mole ratio difference to get the moles, but I'm not sure how. [/QUOTE]
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How do you calculate moles to neutralise oxalic acid?
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