Answer to Question #153738 in General Chemistry for Yusef Silverio

Question

1. calculate the volume O₂ (g) measured at 20^oC and 120.0 KPa that can be obtained from the decomposition of 15.0 g of KNO₃ from the reaction: 2 KNO₃ (s) --> 2 KNO₂ (s) + O₂ (g)
2. How many liters of O₂ (g) measured at 30^oC and 110.0 KPa is needed to completely burn 20.0 liters of butane, C₄H₁₀(g) + 13/2 O₂ (g) -> CO₂(g) + 5 H₂O(/).

Answer 1

1.

number of moles in 15g of KNO3 = 15/101 = 0.149 moles

$2 KNO_{3 (s)} \to 2 KNO_{2 (s)} + O_{2 (g)}$

At STP, 2 moles of KNO₃ decompose to give 1 mole of O₂

This means that 0.149 moles of KNO₃ will decompose to give 0.0745 ($\frac{0.149}{2}$) moles of O₂

T = 20°C = 293K

P = 120.0kPa = 120,000 Pa

n = 0.0745 moles

using the ideal gas law,

PV = nRT

$V = \dfrac{nRT}{P} = \dfrac{0.0745 ×8.314× 293}{120000}$

= 0.00151m³ = 1.51dm³ (1.51L)

2.

20.0 litres of butane = 20/22.4 moles = 0.893 moles

$C_4H_{10(g)} + \frac{13}2 O_{2 (g)} \to CO_{2(g)} + 5 H_2O_{(l)}$

According to the reaction above,

1 mole of C₄H₁₀ uses 6.5 moles of O₂ to combust completely

This means that 0.893 moles of C₄H₁₀ uses 5.8 (0.893×6.5) moles of O₂ to combust completely

T = 30°C = 303K

P = 110.0kPa = 110,000 Pa

n = 0.893 moles

using the ideal gas law,

PV = nRT

$V = \dfrac{nRT}{P} = \dfrac{5.8 ×8.314× 303}{110000}$

= 0.1328m³ = 132.8dm³ = 132.8L