2 years ago

The combustion of hydrogen and oxygen is commonly used to

Chemistry

1 answer:

posledela2 years ago

5 0

Answer:

6.66 mol

Explanation:

(atm x L) ÷ (0.0821 x K)

(0.875 x 250) ÷ (0.0821 x 400)

=6.66108

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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

Dima020 [189]

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.

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3 years ago