In this case, according to the equation for the calculation of the total heat of reaction when a fixed mass of a fuel like ethane is burnt, we can write:

$Q=n*\Delta _cH$

Whereas n stands for the moles and the other term for the enthalpy of combustion. Thus, for the required total heat of reaction, we first compute the moles of ethane in 3 g as shown below:

$n=3g*\frac{1mol}{30.08g}=0.1mol$

Next, we understand that -337.0kcal is the heat released by the combustion of 1 mole of ethane, therefore, to compute Q, we proceed as follows:

$Q=0.1mol*-337.0\frac{kcal}{mol}\\\\Q=-33.6kcal$

Best regards!

8 0

2 years ago

The equilibrium constant for the chemical equation N2(g) + 3H2(g) ⇌ 2NH3(g) and Kp=0.174 at 243°C. Calculate the value of Kc for

Mashcka [7]

Answer:

The Kc of this reaction is 311.97

Explanation:

Step 1: Data given

Kp = 0.174

Temperature = 243 °C

Step 2: The balanced equation

N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 3: Calculate Kc

Kp = Kc *(RT)^Δn

⇒ with Kp = 0.174

⇒ with Kc = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 Latm/Kmol

⇒ with T = the temperature = 243 °C = 516 K

⇒ with Δn = number of moles products - moles reactants 2 – (1 + 3) = -2

0.174 = Kc (0.08206*516)^-2

Kc = 311.97

The Kc of this reaction is 311.97

3 0

3 years ago

What is the ratio of mass of 0.7moles of nitrogen gas to 11.2liters of oxigen gas at STP?​

What is the ratio of mass of 0.7moles of nitrogen gas to 11.2liters of oxigen gas at STP?