Step 4: Substitute Coefficients and Verify Result. Count the number of atoms of each element on each side of the equation and verify that all elements and electrons (if there are charges/ions) are balanced. 3 NaOH + Cr (NO3)3 = 3 NaNO3 + Cr (OH)3. Reactants. Cr2o3+Koh+H2o. In this reaction Potassium hexahydroxochromate (III) is produced which is a complex type of molecule, by balancing the reactant and product side we get, Cr2O3 + 6 KOH + 3 H2O = 2 K3 [Cr (OH)6]. If we check the oxidation state for Chromium (Cr) in the reactant side and in the product side it changes from (3+) to (6+), the 3. The oxidation number of hydrogen is +1 when it is combined with a nonmetal as in CH 4, NH 3, H 2 O, and HCl. 4. The oxidation number of hydrogen is -1 when it is combined with a metal as in. LiH, NaH, CaH 2, and LiAlH 4. 5. The metals in Group IA form compounds (such as Li 3 N and Na 2 S) in which the metal atom has an oxidation number of +1 Chemistry questions and answers. QUESTION #1: Assign an oxidation number to each atom in the following compound: Fe (OH)3Fe (OH)3. Express your answers as signed integers separated by commas. QUESTION #2: Identify the oxidizing and reducing agents. Drag the appropriate labels to their respective targets. Balanced Chemical Equation. 2 Cr (OH) 3 + 3 H 2 O 2 + 4 NaOH → 2 Na 2 CrO 4 + 8 H 2 O. Warning: One of the compounds in Cr (OH)3 + H2O2 + NaOH = Na2CrO4 + H2O is unrecognized. Verify 'Cr (OH)3' is entered correctly. ⬇ Scroll down to see reaction info and a step-by-step answer, or balance another equation. Multiply coefficient for Cr(OH) 3 by 2 1 Cr + 3 Fe(OH) 2 = 2 Cr(OH) 3 + 1 Fe H is balanced: 6 atoms in reagents and 6 atoms in products. Fe is not balanced: 3 atoms in reagents and 1 atom in products. In order to balance Fe on both sides we: Multiply coefficient for Fe by 3 1 Cr + 3 Fe(OH) 2 = 2 Cr(OH) 3 + 3 Fe Cr is not balanced: 1 atom in Balanced Chemical Equation. 2 Cr (OH) 3 + 5 H 2 O 2 → 2 CrO 4 + 8 H 2 O. Warning: 2 of the compounds in Cr (OH)3 + H2O2 = CrO4 + H2O are unrecognized. Verify the equation was entered correctly. ⬇ Scroll down to see reaction info and a step-by-step answer, or balance another equation. Assign oxidation numbers to each element in the following ions. (a) Cr(OH)4-Assign oxidation numbers to each element in the following ions. (f) V2O7 4-Nitrogen can have several different oxidation numbers ranging in value from -3 to +5. (b) Based on oxidation n Nitrogen can have several different oxidation numbers ranging in value from -3 to +5. Le1m.