In the manufacturing process of sulfuric acid, sulfur dioxide is reacted with oxygen to produce sulfur trioxide. using the equation, 2so2 (g) + o2 imported asset 2so3 (g), if 64.06g of sulfur dioxide is given an opportunity to react with an excess of oxygen to produce 75.00 g of sulfur trioxide, what is the percent yield of this reaction? 46.83% 60.25% 75.55% 93.68%

Answer : The percent yield of the reaction is, 93.68 %

Explanation :  Given,

Mass of = 64.06 g

Molar mass of = 64 g/mole

Molar mass of = 80 g/mole

First we have to calculate the moles of .

Now we have to calculate the moles of .

The balanced chemical reaction will be,

From the balanced reaction, we conclude that

As, 2 moles of react to give 2 moles of

So, 1.0009 moles of react to give 1.0009 moles of

Now we have to calculate the mass of

The theoretical yield of  = 80.072 g

The actual yield of  = 75.00 g

Now we have to calculate the percent yield of

Therefore, the percent yield of the reaction is, 93.68 %

option c: 1605.1 kj

the balanced chemical reaction for combustion of methane is as follows:

here, enthalpy formation of is -74.7 kj/mol, 0 kj/mol, -393.5 kj/mol and -241.82 kj/mol respectively.

the heat released or can be calculated as follow:

putting the values,

thus, heat released by combustion of 1 mol of methane is 802.54 kj/mol. thus, for 2 mol, heat released will be:

therefore, heat released by the combustion of 2 mol of methane is 1605.1 kj.