Cho sơ đồ phản ứng sau: X \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbw5usTq % vATv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwz % YbItLDharqqtubsr4rNCHbGeaGqiVGI8Vfeu0dXdh9vqqj-hEeeu0x % Xdbba9frFj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9q8qqaq-dir-f % 0-yqaiVgFr0xfr-xfr-xb9adbaGaaiaadaWaamaaceGaaqaacaqbaa % GcbaWaa4ajaSqaaiaabUcacaqGGaGaaeOtaiaabggacaqGpbGaaeis % aiaabccacaqGOaGaaeiBaiaab+gacaqGHbGaaey9aiaab6gacaqGNb % GaaeilaiaabccacaqGKbGaaeO9aiaabMcaaeqakiaawkziaaaa!4BD4! \xrightarrow{{{\text{ + NaOH (loãng, dư)}}}}\)Y \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbw5usTq % vATv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwz % YbItLDharqqtubsr4rNCHbGeaGqiVGI8Vfeu0dXdh9vqqj-hEeeu0x % Xdbba9frFj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9q8qqaq-dir-f % 0-yqaiVgFr0xfr-xfr-xb9adbaGaaiaadaWaamaaceGaaqaacaqbaa % GcbaWaa4ajaSqaaiaabUcacaqGGaGaaeisamaaBaaameaacaaIYaaa % beaaliaabofacaqGpbWaaSbaaWqaaiaaisdaaeqaaSGaaeiiaiaabI % cacaqGSbGaae4BaiaabggacaqG1dGaaeOBaiaabEgacaqGPaaabeGc % caGLsgcaaaa!492D! \xrightarrow{{{\text{ + }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}{\text{ (loãng)}}}}\)Z \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbw5usTq % vATv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwz % YbItLDharqqtubsr4rNCHbGeaGqiVGI8Vfeu0dXdh9vqqj-hEeeu0x % Xdbba9frFj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9q8qqaq-dir-f % 0-yqaiVgFr0xfr-xfr-xb9adbaGaaiaadaWaamaaceGaaqaacaqbaa % GcbaWaa4ajaSqaaiaabUcacaqGGaGaaeOtaiaabggacaqGjbGaaeii % aiaabUcacaqGGaGaaeisamaaBaaameaacaaIYaaabeaaliaabofaca % qGpbWaaSbaaWqaaiaaisdaaeqaaSGaaeiiaiaabIcacaqGSbGaae4B % aiaabggacaqG1dGaaeOBaiaabEgacaqGPaaabeGccaGLsgcaaaa!4DA2! \xrightarrow{{{\text{ + NaI + }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}{\text{ (loãng)}}}}\) T.
Biết rằng X, Y, Z, T là các hợp chất của crom. Hai chất X và Z lần lượt là:
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Lời giải:
Báo saiCrO3 + 2NaOH → Na2CrO4 + H2O
Na2CrO4 + H2SO4 → Na2Cr2O7 + Na2SO4 + H2O
Na2Cr2O7 + NaI + H2SO4 → Cr2(SO4)3 + I2 + Na2SO4 + H2O