Answer :
Let's find out how many grams of sulfur would be left over if 10 grams of lead (Pb) were mixed with 5 grams of sulfur (S).
1. Determine the moles of lead (Pb):
- The atomic weight of lead (Pb) is 207.2 grams per mole.
- Moles of lead [tex]\( = \frac{ \text{mass of lead} }{ \text{atomic weight of lead} } = \frac{10 \text{ grams}}{207.2 \text{ grams/mole}} \)[/tex]
- Moles of lead [tex]\( = 0.04826254826254826 \)[/tex].
2. Determine the moles of sulfur (S):
- The atomic weight of sulfur (S) is 32.06 grams per mole.
- Moles of sulfur [tex]\( = \frac{ \text{mass of sulfur} }{ \text{atomic weight of sulfur} } = \frac{5 \text{ grams}}{32.06 \text{ grams/mole}} \)[/tex]
- Moles of sulfur [tex]\( = 0.15595757953836556 \)[/tex].
3. Check the reaction ratio:
- Lead (II) sulfide (PbS) is formed in a 1:1 molar ratio of lead to sulfur. This means one mole of lead reacts with one mole of sulfur.
4. Determine the limiting reagent:
- Compare the moles of lead and the moles of sulfur.
- Since there are fewer moles of lead (0.04826254826254826) compared to sulfur (0.15595757953836556), lead (Pb) is the limiting reagent. Therefore, only 0.04826254826254826 moles of sulfur will react with the available lead.
5. Calculate the leftover sulfur:
- Initially, there are 0.15595757953836556 moles of sulfur.
- Moles of sulfur that react with lead [tex]\( = 0.04826254826254826 \)[/tex].
- Moles of sulfur left over [tex]\( = 0.15595757953836556 - 0.04826254826254826 = 0.1076950312758173 \)[/tex].
6. Convert the leftover moles of sulfur to grams:
- Mass of leftover sulfur [tex]\( = \text{moles of leftover sulfur} \times \text{atomic weight of sulfur} \)[/tex]
- Mass of leftover sulfur [tex]\( = 0.1076950312758173 \times 32.06 \text{ grams/mole} = 3.4527027027027026 \text{ grams} \)[/tex].
Thus, if 10 grams of lead is mixed with 5 grams of sulfur, approximately 3.45 grams of sulfur would be left over.
1. Determine the moles of lead (Pb):
- The atomic weight of lead (Pb) is 207.2 grams per mole.
- Moles of lead [tex]\( = \frac{ \text{mass of lead} }{ \text{atomic weight of lead} } = \frac{10 \text{ grams}}{207.2 \text{ grams/mole}} \)[/tex]
- Moles of lead [tex]\( = 0.04826254826254826 \)[/tex].
2. Determine the moles of sulfur (S):
- The atomic weight of sulfur (S) is 32.06 grams per mole.
- Moles of sulfur [tex]\( = \frac{ \text{mass of sulfur} }{ \text{atomic weight of sulfur} } = \frac{5 \text{ grams}}{32.06 \text{ grams/mole}} \)[/tex]
- Moles of sulfur [tex]\( = 0.15595757953836556 \)[/tex].
3. Check the reaction ratio:
- Lead (II) sulfide (PbS) is formed in a 1:1 molar ratio of lead to sulfur. This means one mole of lead reacts with one mole of sulfur.
4. Determine the limiting reagent:
- Compare the moles of lead and the moles of sulfur.
- Since there are fewer moles of lead (0.04826254826254826) compared to sulfur (0.15595757953836556), lead (Pb) is the limiting reagent. Therefore, only 0.04826254826254826 moles of sulfur will react with the available lead.
5. Calculate the leftover sulfur:
- Initially, there are 0.15595757953836556 moles of sulfur.
- Moles of sulfur that react with lead [tex]\( = 0.04826254826254826 \)[/tex].
- Moles of sulfur left over [tex]\( = 0.15595757953836556 - 0.04826254826254826 = 0.1076950312758173 \)[/tex].
6. Convert the leftover moles of sulfur to grams:
- Mass of leftover sulfur [tex]\( = \text{moles of leftover sulfur} \times \text{atomic weight of sulfur} \)[/tex]
- Mass of leftover sulfur [tex]\( = 0.1076950312758173 \times 32.06 \text{ grams/mole} = 3.4527027027027026 \text{ grams} \)[/tex].
Thus, if 10 grams of lead is mixed with 5 grams of sulfur, approximately 3.45 grams of sulfur would be left over.
