Areaction vessel contains 10.0 g of co and 10.0 g of o2. how many grams of co2 could be produced according to the following reaction? 2 co(g) + o2(g) → 2 co2(g) calculate the efficiency of the vessel if 6 g of co2(g) was collected after the process.

1. 15.71 g CO2

2. 38.19 % of efficiency

Explanation:

According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:

For CO:

C = 12.01 g/mol

O = 16 g/mol

CO = 28.01 g/mol

(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO

For CO2:

C = 12.01 g/mol

O = 16 x 2 = 32 g/mol

CO2 = 44.01 g/mol

We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:

(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2

We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:

(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2

Now for the efficiency question:

From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,

(6g / 15.71g) x 100 = 38.19 % of efficiency

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.