Question

In the reaction, 2 HgO(s) → 2 Hg(s) + 0,() what mass of mercury(II) oxide, HgO, would be required to produce 519 L of oxygen, O, measured at STP? mass:

Answer 1

22.4 L OR I mole of gas. I moles of gas. 22.4 Volume Istock H₂O₂ required to make diluted Solution Q.Nol: The given equation is argo 24g + O2 Volume of oe needed = 5191 At STP, I mole of gas 22.4 L IL 5192 x 519 22.4 23.17 moles of gas i.e., Moles of O2 - 23.17 mol's from the reaction above, it is clear that 2 moles of rgo as required to produce 02 - il moles of Hgo required to produce 23.17 moles of Oz I* 23.17 = 11:58 moles : 216.59 glomol. .: Amount of rgo required to produced 519L of Cz Moles 4. go * Molenr mais of Ugo = 11:58x216:59 09.199 4.9M P.No:2: Molarity of stock 420, - = M, Molarity diluted H2O2 = 0.80M M2 diluted H₂O₂ = 50.0mL = V₂ Volume of H₂O₂ solution = ? Imole. 2 Molor mars of rego = 2509.199 82.51 kg f

Using Molosity Equation, we have MV, M2 49xV = 0.80 X 50 8.16 mL Nolume Q. No. 3: Molarity of Noon solution = 0.300 M Naon solution = 247ml = 0.2472 Now, Molasity is given by Molasity No. of moles volume of solution No. of moles Noon = Molarity of Naon x volume of - 0.074) moles Now, Molar mars of NaOH = 40 40 g/mol. .. Amount of Noon requireal Noonsdution こ 0.300x0.247 of Maon - Moles Naon * Molar mars 0.0741 x 40 = 2.96 of g me to answer the Note: It is not possible for remaining questions because the mage last is not opening up.