Answer to Question #159878 in General Chemistry for jordan

Question

1.) The  head  of  a  match  contains  approximately  0.75 0 g  of  diphosphorus trisulfide. When  the  match  is  struck  on  a  rough  surface , it combusts  and explodes into  a flame  producing  diphosphorus  pentaoxide  and  sulfur  dioxide. What is  the percent  yield of  sulfur  dioxide if  0.2895  L  is produced when 0.88 g  of  oxygen is available at  a temperature  is  26.5°C and a pressure of 14.9 kPa?  

2.)

Ethanol (C2H6OH)  vapour  completely  combusts  in  air. If  2.5  L  of  ethanol burns  at 1.4  atm  and  18.6  ⁰C, what  volume  of  oxygen  is  required?  What  mass  of  carbon dioxide  will  be  produced? 

Answer 1

1. )P₂S₃ + 11/2 O₂$\rightarrow$ P₂O₅ + 3SO₂

Amount of SO₂ produced :

PV = nRT

n = 0.147 ×0.2895/ (0.0821×299.5)

n = 0.0017 moles

Theoretical yield in moles = 3 x moles of P₂S₃

= 3 × (0.75/158) = 0.0142 moles

%yield = (0.0017/0.0142)x100 = 11.97%

2.) 2C₂H₅ +13/2 O₂$\rightarrow$4 CO₂ + 5H₂O

PV = nRT

n = 1.4 x 2.5 /(0.0821×291.6)

n = 0.143moles of ethanol

So no. Of moles of oxygen required =

(13/4) × 0.143 = 0.4655 moles

V = nRT/P

V = 0.4655 x 0.0821 ×291.6 /1.4 = 7.96 L

No. Of moles of Carbondioxide produced =

2×0.143 = 0.286 moles

Mass = 0.286x44 = 12.584 g