Question

7. How many Oxygen atoms are there in 29 gen atoms are there in 296.0 g of Cr(PO4)2 ? show work (6 pts) consider the following balanced chemical equation when answering the following questions: 2 Cs3PO3 + 3 BaBr2 ------ ----→ 6 CsBr + Ba3(PO3)2 a) How many mols of CsBr should be produced if 8.68 g of BaBr2 are used in the reaction? SHOWWORK (5 pts) 9g BaBrzx Imole Bobr? b) If 92.7 g of Cs3PO3 are reacted with 0.463 mol of BaBr2, then how many mol of CsBr will be produced? SHOW WORK (10 pts)

1) How many oxygen atoms are there in 296.0 g of Cr(PO4)2

2) ?
a)
b)

Answer 1

7. Mole of Cr (PO4)2 = 296.0g 1 242 gl mol - 1.223 mol = 0.0292 mol mol Mole of Onygen = 8 x 1.223 z 9.785 mol Numbes of Oxygen = 6:02 x 1023 x 9.785 = 58.91 x 1023 atomi = 15.89 x 1024 /atom of oxygen 8 @ Mole og BaBra = 8.684 m = 0.0292 mol 2 978 Imol . Mole of Cs Br = 2x0.0292 = 0.0584 mol - 89 of BABY ₂ x 1 mol Babra x 6 mol of Co Br : 0.0584 1 2979 Babra . 3 mol of Bar, o 6) 2Cspoz + 3 BaBra — 6 Cs Br + BazPez)z 92.79 0.463 mot 470.7glmol 0.194 mol limiting reagent Mole of Cs Br = 3x Mole of CS₂ P0₃ E 3X0.194 mol = 10.582 mol