Toluene (C_7H_8) and nitric acid (HNO_3) are used in the production of trinitrot

Question

Toluene (C_7H_8) and nitric acid (HNO_3) are used in the production of trinitrotoluene (TNT, C_7H_5N_3O_6), an explosive. C_7H_8 + HNO_3 rightarrow C_7H_5N_3O_6 + H_2O (not balanced) (a) what mass in grams of nitric acid is required to react with 454 g C_7H_8 (b) what mass in grams of TNT can be made from 829 g C_7H_8? Describe how you would prepare 500.0 mL of an aqueous solution that is 30.0% isopropyl alcohol by volume. Please answer the two questions below, after reading Chapter 6 in the textbook and reviewing the correspond Powerpoint slides. Make sure that you give your answer in a clear and justified manner, providing necessary explanation on how did you come up with it. Please post your initial thread by Thur May 25th, 11:59pm. Also, please reply to the posts of at least 3 peers by Sun May 28th. In order to reply, click on the peer's thread first, then click reply. Calculate the density, in grams per liter, of carbon dioxide (CO_2) gas at STP. How many grams of CO(g) are there in 74.5 mL of the gas at 0.933 atm and 30 degree C?

Explanation / Answer

ANSWER

1 A) =930 gm because one mole toluene react with three mole of nitiric acid and here 454 gm have 4.9 mole so nitric acid have 4.9x3 mole and weight =molex molar mass C7H8 + 3 HNO3 = C7H5N3O6 + 3 H2O

b) ==2044 gm because 829gm/92.14[molar mass toluene] =8.99 mole so equal to tnt cause 8.99x227.13[molar mass tnt] =2044 gm

2..   

We can use the Ideal Gas Law. Let's assume that we have 1 mol of CO.

PV=nRT

V=nRTP=1 mol×8.314 kPaLK1mol1×273.15 K100 kPa = 22.71 L

And 1 mol CO2×44.01 g CO21 mol CO2=44.01 g CO2

Density = massvolume=44.01 g22.71 L = 1.938 g/L

b) PV =NRT

0.933X0.0745=NX0.082X303

NOW N =0.933X0.0745/0.082X303 =0.027 MOLE

WEIGHT =MOLE X MOLER MASS =0.027 X28 =0.756 gm

3

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