Question

27. 1/2 points Previous Answers ncchemap 6.041. CS (0) +0,0) + CO2(0) + S02() (a) Consider the unbalanced equation above. What is the maximum mass of so, that can be produced when 36.0 g of cs, and 64.5 g 0, react? 40. 60.6 g Soz (b) Assume the reaction in part (a) goes to completion. Calculate the mass of excess reactant. 40) 25.5 X 9 28. + 1/2 points Previous Answers ncchemap 6.044. N, (.) +0,6) ► N2050) (a) Consider the unbalanced equation above. What is the maximum mass of N, O, that can be produced when 25.5 g of N, and 29.0 g 0, react? 40 39.2 ONZO (b) Assume the reaction in part (a) goes to completion. Calculate the mass of excess reactant. 49 10.2 Submit Answer Practice Another Version

Answer 1

27)                        CS₂    + 3O₂    -------> CO₂   +   2SO₂

Molecular mass: 12+2*32        3*16                          32+ 2*16

                      = 76g              = 96g                         =128g

Now, 76g of CS₂ require 96g of O₂

36g of CS₂ require = (96* 36)/76 = 45.6 g of O₂

Oxygen available =64.5g

oxygen required= 45.67g

therefore, oxygen excess = 64.5 - 45.47 =19.03g

Now,

76g of CS₂ produces 128g of SO₂

so, 36g of CS₂ produces (128*36)/76 = 60.6 g of SO₂

(a) Maximum mass of SO₂ produced when given amount of reactants are adding is 60.6g

(b) Mass of excess reactant(oxygen) = 19.03g

28) 2N₂ + 5O₂ ------> 2N₂O₅

molecular mass:    28*2     32*5         2*(2*14 + 5*16)

                           =56g     =25.5g      =216g

now, 160g of oxygen require 56g of nitrogen

        29g of oxygen require (56* 29)/160 = 10.15g of nitrogen

N₂ available = 25.5 g

N₂ reacting= 10.15g

N₂ excess = 25.5-10.15 = 15.35g

Now,

160g of O₂ reacts to give 216g of N₂O₅

29g O₂ reacts to give (216*29)/160 = 39.15g of N₂O₅

(a) Maximum mass of N₂O₅ that can be produced from given amount of reactants is 39.15g

(b) Mass of excess reactant (dinitrogen) = 15.35g