Question

2) 4NH3(g) + 302(g) + 6H2O(l) + 2Nz(9) (a) 2 moles of NH, and 3 moles of Oz will produce ? moles Nz? (b) 5.0 g of NH3 and 5.0 g of O, will produce ? g of H:O? How many H2O molecules? 3) A mixture of 80.0 g of chromium (VI) oxide, Cr:Os, and 8.00 g of solid carbon, C, is used to produce elemental chromium, Cr, by the reaction below Cros (aq) + 3C (s) → 2Cr (s) + 3C0 (9) a) Which reactant is the limiting reagent? b) What is the theoretical yield of Cr that can be obtained from the reaction mixture?

Answer 1

H 4 NH3 gi + 302cg, 6H2O(l) + 2 H21g) according to above equation, a moles of MHz egy react with 3 moles of Orego will give 6 moles of H2O & 2 moles of Mr. . as. gium moles of Mirz = 2 " " of 02-3 (a) so, Nilz is limiting reagent and will give product 1 yield. . This , as u moles of NH3 gives a moles of Na 1 So, a moles of NH3 will give 1 mole of N2. Thiss 2 moles of NH3 & 3 moles of Oz will produce - 1 mole of NHz (b) moles of Mility in sg - Samol = 0.294 mole moles of Oz in 59 = 59 . 32gl mol = 0.156 mol 4 moles of NH3 react with 3 mole of 0₂ so. 0.294 mole of NH3 react with 3 x 0.294 0.2205 mole of but On present is 0.156 mol - This or is limiting reagent and will give product as 3 mole of a give b moles of Thus i 156 mole of 02 give 0.312 marts This mass of H2O in 0.312 mole 17 9 mar yield: ole of Oz give 0.312 mole of H2O H20 in 0.312 mole = 5.610g of ho

Thus Jg of O₂ will produce 5.616 g of H₂O. I mole= 6.022 x10 23 molecule 0.312 mole of H2O = 6,022x10 23 x 0.312 molecule of tho 0 - 11 mol e x H, O = 1. 818 8 loº 3 mol alc dự Hộp will be produced . 80g = 0.526 mole - (3) molar mass of C%2O₂ = 152 g I mol molar mass of (z = 12 glmal maar mass of lr 52 glmal molau mass og co = 28 gimal CO2O3(aq) + 3053) + 2Cr(s) + 31069) (1) mole of Co2 Oz in so g = 8o a me mole of c in 8 g = a mai 0.67 mole i mole of Cr2O₃ react with 3 mole of 0.526 mole g (7203 will leact with 3x0526 mole = 1.578 m But mole ofc is 0.67 (presents fC. so, cis limiting reagent. (b) 3 more of a give a mole of co This 0.67 mole of c will give 2x0.67 more oro de X0:67 mole of Co mase of cr obtained = o.us mole x 52 g/mole = 0.45 mole - 23.22 g of co Thus 23.22 g g co will obtain

(0) yield: - X 100 actual yield Theo ritical yield X 100 - 93.42% Yo yield > 21.7_9 - 23.22 g I WIL