Q1)hydrogen peroxide H2O2 reacts with magnet (v i i) MnO4⁻ in basic solution according to following equation 2MnO4⁻(aq)+3H2O2(aq),>>>2MnO2 (S)+3O₂( g)+2OH-(aq)+2H2O(l) How many moles of hydrogen peroxide would be needed to produce eight moles of water (answer is 12) please show the workings and formula used.
Q2 ) 20 gm of Zinc reacted with 20 gm of Iodine.Calculte the mass of Zinc and Iodine.Which is excess reagent?
Q3)Aqueous XO4^-3 ion forms a precipitate with aqueous Ag ion.With a concentration of 0.2 mole/dm^-3 is added to a solution of ions1.2 gram of precipitate is formed Calculate the mole of silver ion used and calculate the mole of Precipitate formed Find Molar mass of Precipitate Determine the RAM of X and identify X .
Sorry I forgot to add one thing:
Q3)Aqueous XO4^-3 ion forms a precipitate with aqueous Ag ion.When 41 cm^3 of solution of silver aqueous ions were added to the solution then 1.2 gram of precipitate is formed
a) Calculate the mole of silver ion used and
b) calculate the amount of mole of Precipitate formed
c)Find Molar mass of Precipitate d)Determine the RAM of X and identify X .
- hcbiochemLv 74 weeks agoFavorite Answer
Q1) The coefficients of the balanced equation allow you to relate moles of one thing to moles of another.
8.0 moles H2O X (3 mol H2O2 / 2 mol H2O) = 12 mol H2O2
Q2) The balanced equation is: Zn + I2 --> ZnI2
From this, you see that 1 mol of Zn reacts with 1 mol I2.
20 g Zn / 65.39 g/mol = 0.306 mol Zn
20 g I2 / 254 g/mol = 0.079 mol I2
Since you have fewer moles of I2, I2 would be the limiting reactant and Zn is present in excess.
Q3) I think you have left out some critical information from this question. I would suggest you just create a new question with all of the information for this one.