A 100 g of ice (solid water) at 0 degrees Celsius melted to water at 75.5 degree

Question

A 100 g of ice (solid water) at 0 degrees Celsius melted to water at 75.5 degrees Celsius. If all the heat came from propane (C3H8) combustion, how many grams of propane (molar mass is 44.1 g/mol) must be combusted? (change of H = 6.02 kj/mol, SH2O (l) =4.187 kj/kg)
Equation: C3H8 (g) + 5 O2 (g) --> 3 CO2 (g) + 4 H2O (l) change Hc = -2202 kj
(When I wrote change, I meant the triangle sign) A 100 g of ice (solid water) at 0 degrees Celsius melted to water at 75.5 degrees Celsius. If all the heat came from propane (C3H8) combustion, how many grams of propane (molar mass is 44.1 g/mol) must be combusted? (change of H = 6.02 kj/mol, SH2O (l) =4.187 kj/kg)
Equation: C3H8 (g) + 5 O2 (g) --> 3 CO2 (g) + 4 H2O (l) change Hc = -2202 kj
(When I wrote change, I meant the triangle sign)
Equation: C3H8 (g) + 5 O2 (g) --> 3 CO2 (g) + 4 H2O (l) change Hc = -2202 kj
(When I wrote change, I meant the triangle sign)

Explanation / Answer

C3H8 (g) + 5 O2 (g) --> 3 CO2 (g) + 4 H2O (l) change Hc = -2202 kj

delta H for the above reaction x number of moles of C3H8 = delta H observed for 100 g of ice = 6.02 kJ/mol

2202 kJ x (wC3H8/44.1) = 6.02 kJ

Now solve for weight of C3H8

wC3H8 = (6.02 x 44.1) / 2202

= 0.1205 g

= 1.2 x 10^-4 kg
  

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