The value of n in $ AlF_XO_Y^n $ , if x=1 and y=1 ?
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Chemistry MCQs for NEET — Practice Questions with Answers
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The value of n in $ AlF_XO_Y ^n$ , if x=2 and y=3
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What will be the value of x and y respectively in $ AlF_XOy^{6–} ? $
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How many moles of elements are added when 2.5 mole $ Cr_2O_7 ^{2–} reduced in Cr^{3+} $ ?
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Explanation
When $Cr_2O_7^{2-}$ is reduced to $Cr^{3+}$, the half-reaction is: $$Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O$$. For 2.5 moles of $Cr_2O_7^{2-}$, the moles of electrons added can be calculated by multiplying 2.5 by the number of electrons involved per mole of $Cr_2O_7^{2-}$: $$2.5 imes 6 = 15 ext{ moles of electrons}$$.
What moles of $Cr_2O_7 ^{2–} reduced in Cr^{3+} $ by the addition of 12 moles of electrons ?
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Explanation
Using the same half-reaction: $$Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O$$. To find the moles of $Cr_2O_7^{2-}$ reduced by 12 moles of electrons, we set up the proportion: $$ rac{12 ext{ moles of } e^-}{6 ext{ moles of } e^-} = 2 ext{ moles of } Cr_2O_7^{2-}$$.
How many mole ferrous $(Fe^{2+}) ion oxidized in ferric (Fe^{3+}) ion by the required no. of electrons the oretically to reduced 4 mole Cr_2O_7^{ 2– }in to Cr^{3+} $ ?
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Explanation
First, determine the total number of moles of electrons needed to reduce 4 moles of $Cr_2O_7^{2-}$ using the same half-reaction: $$4 imes 6 = 24 ext{ moles of electrons}$$. Since each $Fe^{2+}$ ion loses one electron to become $Fe^{3+}$, 24 moles of electrons will oxidize 24 moles of $Fe^{2+}$.
What mole of $MnO_4 ^– reduced in Mn^{2+} by the addition of 7.5 mole electrons in MnO_4 ^ –$ ?
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How many electrons required to add for the reduction of one mole of $MnO_4 ^ – in Mn^{2+} $ ?
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When $ 3.11 \times 10^{24} Cr_2O_7 ^{2–} ion reduced in Cr^{3+}, then how many ferrous (Fe^{2+}) ionoxidised in ferric (Fe^{3+}) $ ion ?
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Theoretically, how many moles of iodide $ (I^–) ion oxidized into iodate (IO_3 ^ –) in using the no of electrons required for the reduction of 24 moles of MnO_4 ^ – ion into Mn^{2+} $ ion ?
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