Question

How many grams of propane are in 20 pounds of propane? Use the conversion 1 lb = 454 g. (Express your answers for the next three questions in scientific notation. For example use 2.3e-5 to indicate a number such as 2.3 x 10-5.) 9.1e3 grams b) How many moles of propane are in 20 pounds of propane? 2.1e2 moles c)How much heat can be obtained by burning 20 pounds of propane? (Remember to look at this from the viewpoint of the surroundings, since the question asks how much heat can be OBTAINED.) 4.6e5 kJ

Answer 1

Answer:

a. $=9.1x10^3gC_3H_8$

b. $2.1x10^2molC_3H_8$

c. $Q=-4.6x10^5kJ$

Explanation:

Hello,

a. By applying the given information, one obtains:

$20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8} =9.1x10^3gC_3H_8$

b. By knowing that the propane has a molecular mass of 44g/mol, one obtains:

$20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8}*\frac{1molC_3H_8}{44gC_3H_8} =2.1x10^2molC_3H_8$

c. Here, the propane's combustion chemical reaction is stated:

$C_3H_8+5O_2-->3CO_2+4H_2O$

This enthalpy of reaction is computed via:

Δ$rH=(3*-393.5kJ/mol)+(4*241kJ/mol)-(-104.7kJ/mol)=-2043kJ/mol$

Finally, since it is done for 20 lb of propane ($2.1x10^2molC_3H_8$), the obtainable energy is:

$Q=-2043kJ/mol*2.1x10^2molC_3H_8\\Q=-4.6x10^5kJ$

Best regards.