When octane (c8h18) is burned in a particular internal combustion engine, the yield of carbon dioxide is 93%. what mass of carbon dioxide will be produced in this engine when 15.0 g of octane (mw = 114.0 g/mol) is burned with 15.0 g of oxygen gas (mw = 32.0 g/mol)? 2c8h18 + 25 o2 --> 16 co2 + 18 h2o
a.12g b.13g c.21g d.43g e.54g

The correct answer is option a.

Explanation:

Moles of octane =

Moles of oxygen gas =

According to reaction, 25 moles of oxygen reacts with 2 moles of octane.

Then 0.46875 moles of oxygen will react with:

of octane

Oxygen is present in limited amount. Hence, limiting reagent.

According to reaction , 25 moles of oxygen gives 16 moles of carbon-dioxide.

Then 0.46875 moles of oxygen will give:

of carbon-dioxide

Mass of 0.3 moles of carbon-dioxide:

0.3 g × 44 g/mol = 13.2 g

Theoretical yield = 13.2 g

Experimental yield = x

Percentage yield of carbon dioxide = 93 %

x =12.27 g ≈ 12 g

12 grams of carbon dioxide will be produced in this engine .

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.