Video Transcript
Carbon tetrachloride can be synthesized by reacting chlorine with methane according to the equation, CH4 plus 4Cl2 reacting to give CCl4 plus 4HCl. If 8.0 grams of HCl is produced, what mass of methane is consumed?
Before we begin, let’s identify the substances this question is talking about. Carbon tetrachloride is CCl4. And this is a colorless liquid under standard conditions. It is a highly volatile liquid. And it’s also a known carcinogen. Chlorine has the formula Cl2 and is a gas under standard conditions. Methane is CH4. And this is the simplest hydrocarbon gas. And HCl is hydrogen chloride, a gas under standard conditions. The equation given to us is already balanced. This reaction is an example of a substitution reaction. And it needs the energy from UV light to occur.
If we write the equation again, we can then write underneath each substance the data that we know for that particular substance. We are told that 8.0 grams of HCl is produced. So we can write the mass of HCl as 8.0 grams underneath the formula for HCl. And we are asked to find what mass of methane is consumed or used up in the reaction. So under CH4, we could write 𝑚 equals question mark. This is the only data given to us. But how will we go from the data that we know to data that we don’t know? We can only relate the data from two different substances using the number of moles and the stochiometric coefficients from the balanced equation.
So for HCl, we will need to calculate its molar mass and then, using its mass and molar mass, determine its number of moles. Then we can relate HCl to methane using the stochiometric coefficients one as to four from the balanced equation. The number of moles of methane that are consumed will be a quarter of the moles of HCl that is produced. Or we could say that the number of moles of HCl produced is four times greater than the number of moles of methane that is needed. We can then determine the molar mass of methane and use the number of moles of methane and its molar mass to determine the mass of methane consumed.
Now that we know the method we’re going to follow, let’s begin our calculations. First, we calculate the molar mass of hydrogen chloride by taking hydrogen’s molar mass value from the periodic table, 1.008, plus chlorine’s value of 35.45. And we get 36.458 grams for every mole of hydrogen chloride. We can now calculate the number of moles of hydrogen chloride by taking its mass divided by its molar mass. Putting in our values, we get 8.0 grams divided by 36.458 grams per mole. And we get 0.219 moles of HCl produced from the reaction.
Let’s now use the number of moles of HCl to determine the number of moles of methane consumed. We know from the equation the number of moles of HCl produced is four times greater than the number of moles of methane consumed. And we can take the number of moles of HCl we have just calculated to find 𝑥. The number of moles of methane consumed. We can solve for 𝑥 by taking 0.219 divided by four. And we get 0.0548 moles of methane consumed.
Let’s clear some space to finish this calculation and let’s write in this last piece of data under methane. Next, let’s calculate methane’s molar mass. Carbon’s molar mass value is 12.011. And there are four hydrogens with a value of 1.008 each, which gives 16.043 grams for every mole of methane. Finally, we can calculate the mass of methane consumed by taking its number of moles timesed by its molar mass. This is the same formula or equation we used previously but rearranged to give mass as the subject of the formula. We can put in our number of moles and molar mass value that we calculated and 0.879 grams is the mass of methane that we get for our answer.
The question has not specified how many decimal places or significant figures we need to report our answer in. But we were given a mass of HCl of 8.0, which is two significant figures. So let’s give our final answer to two significant figures. And in this case, it’s 0.88 grams of methane. Finally, if 8.0 grams of HCL is produced, the mass of methane that is consumed is 0.88 grams.
