Question

ONLY need help with C out of part 6, and need help with B out of Part 4. Thank you! ☺️

H2O _Mg(OH)2 + __HPO4 → _Mg 3(PO4)2 + In the reaction above (Mg = 24.31; O = 16.00; H = 1.01; P = 30.97) a) 3.498 x 100 g of H Po, will require how many moles of Mg(OH)2? b) 3.498 x 10° g of H3PO, will produce how many molecules of Mg 3(PO4)2? c) 3.498 x 100 g of H2PO4 will require how many grams of H,0? d) 9.876 x 1025 molecules of H 0 will need how many grams of Mg(OH),? e) 9.876 x 10 molecules of H20 will need how many molecules of H3PO4? 1) 9.876 x 1040 molecules of H20 and how many moles of Mgz(PC), will be made together?

Answer 1

6.c 3 Mg(OH)₂ + 2H₂P04 mg (pay)2 + 6H₂O Molar man looglmol 18 g/mol According to Stoichiometry. 200g 108g (2xloo) (6x18) . 2oog of H₂poy gives logg of H2O & .. 3.498 xlob g " " " 108gx 3. 498 xlog Note: It is not REQUIRE X 1.89x106g of tho It should be PRODUCE ✓ 2oog 4.6 Element Mass Atomic mass (g/mol) Moles (mass/atomic Simplest Mole. number reato Mass Ο 12 (5.93/12) 0.4941 Ι 5.93g 0.4559 2.30g 1.315g 0.455 6.0126 0.44 5.53 0.0822 1.9962 [0.1 643 ] 0.455 2 14 16 0.1643 0.0822 Ο -0.0822] for calculating Simplel mole number ratio, man no. of moles of smallest at element is divided with everyone. o for H it is coming at 5053 we can not approximate anything We will multiply each moles with 2 : C = 12 1 Empirical formula = C12H, NO H 5.5X2 → 11 Na 2x2 = 4 Thus the empirical formula was Da 2 Calculated doing elemental analyeis. 121) 42