How many liters of antimony (III) chloride can be produced from 35 L of chlorine at STP reacting with an excess of antimony? 2Sb + 3Cl2 → 2SbCl3

Answer 1

Note: There is a flaw in this question. In the end, we're supposed to have a volume of SbCl₃ in liters; however, SbCl₃ is a solid. We can still determine the amount of SbCl₃ produced in the reaction, but we'll have to report it as a mass instead of as a volume.

Let's start by identifying the limiting reagent in the reaction. As stated, "an excess of antimony" is reacted with chlorine gas. Since antimony (Sb) is in excess, chlorine gas (Cl₂) is the limiting reagent. This means that the maximum amount of the product SbCl₃ which can be produced is equal to the moles of chlorine gas reacted.

Next, we should convert the volume of chlorine gas to moles. For this, we need to know that one mole of gas occupies 22.4 liters at standard temperature and pressure (STP). So, we divide the given volume of Cl₂ by this value: (35 L Cl₂)/(22.4 L/mol) = 1.5625 mol Cl₂

Now that we know how many moles of Cl2 we start with, we look at the balanced chemical equation and observe that for every three moles of Cl2 consumed, we can produce two moles of SbCl3. So, we should multiply by 2/3: (1.5625 mol Cl₂)*(2 mol SbCl₃)/(3 mol Cl₂) = 1.0416667 mol SbCl₃

Finally, we calculate the mass of SbCl₃ which can be produced by multiplying the number of moles of SbCl₃ by the molecular weight of the compound, which is 228.11 g/mol (and can be obtained by adding the masses of one Sb atom and three Cl atoms): (1.0416667 mol SbCl₃)*(228.11 g/mol) = 237.6156 g SbCl₃

Considering the number of significant figures provided to us (two, since we started with 35 L of Cl₂), the answer should be stated as 240 g SbCl₃.