Question

described in (a). 11. Joseph Priestley's study of the decomposition of mercury(II) oxide (HgO) with heat led to the discovery of O2. The balanced equation for this reaction is shown in Equation 9. (Eq. 9) beat 2 HgO(s) 2 Hg()+O2(g) (a) How many grams of O2 could possibly be formed by the decomposition of 75.0 g of HgO? (b) How many grams of HgO must be decomposed to produce 75.0 g of O2? 12. Propane (CHs) is used as a fuel in many gas grills. The balanced equation for the combustion of CHg is shown in Equation 10. C3Hs(g)+502(g) 3CO2 (g) +4H2O(g) +energy (Eq. 10) (a) How many grams of CO2 could possibly be produced if 10.0 g of C3Hg reacts with an ample supply of O2? (b) How many grams of CO2 could possibly be produced with an ample supply of CaHs, but only 10.0 g of O2? (c) If this reaction is run using 10.0 g of CHg and 10.0 g of O2, what mass of CO2 could possibly be produced? (d) Identify the limiting reagent in the reaction described in (c) 13. Tetraphosphorus decaoxide (P4O10) reacts with H2O to form HAPO4 The balanced equation for this reaction is shown in Equation 11. P&Oio(s)+6H20(e)-4H3PO 4 (e) (Eq. 11) Calculate the mass of H3PO4 that could possibly be produced by the reaction of 300g of P4O10 with 100 g of H2O. 14. Copper (Cu) dissolves in nitric acid (HNO3), producing copper(II) nitrate, Cu(NOs)2 nitrogen dioxide (NO2), and H2O. The balanced equation for this reaction is shown in Equation 12 Cu(s)+4HNO3(aq) Cu(NOs)2(aq)+ 2 NO2(g) +2 H20( ) (Eq. 12)

Answer 1

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heat 4 1 2490 is molar mu 216.59 2 Hg (2) 200159 0. 19) 32.0 | /Imo و 6) al I moles of Hgo produce I mole of 02. 2x216.59 g of Hgo foom 320g of O2. → 75.0g of Hgo form O2 = 32.09 x 75.09 &x 216.59 g = 5.549 - Hence, 5.54 g of 2 will form b) I mole of O2 produce by a moles of Hgo 320g of 02 produce by Q x 216.59 g of Hgo. => 75.0g of oz boucluce by Hg 0 = 2 X216.82g 75.09 320g = 1015 g Hence , 1015 g of HO must be decomposed