Answer:
![\boxed {\boxed {\sf About \ 2.127 \ moles \ of \ FeCl_3}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%20About%20%5C%202.127%20%5C%20moles%20%5C%20of%20%5C%20FeCl_3%7D%7D)
Explanation:
To convert from moles to grams, the molar mass must be used.
1. Find Molar Mass
The compound is iron (III) chloride: FeCl₃
First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).
There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.
- FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol
This number tells us the grams of FeCl₃ in 1 mole.
2. Calculate Moles
Use the number as a ratio.
![\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}](https://tex.z-dn.net/?f=%5Cfrac%7B162.19%20%5C%20g%20%5C%20FeCl_3%7D%7B1%20%5C%20mol%20%5C%20FeCl_3%7D)
Multiply by the given number of grams, 345.0.
![345.0 \ g \ FeCl_3 *\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}](https://tex.z-dn.net/?f=345.0%20%5C%20g%20%5C%20FeCl_3%20%2A%5Cfrac%7B162.19%20%5C%20g%20%5C%20FeCl_3%7D%7B1%20%5C%20mol%20%5C%20FeCl_3%7D)
Flip the fraction so the grams of FeCl₃ will cancel.
![345.0 \ g \ FeCl_3 *\frac{1 \ mol \ FeCl_3}{162.19 \ g \ FeCl_3}](https://tex.z-dn.net/?f=345.0%20%5C%20g%20%5C%20FeCl_3%20%2A%5Cfrac%7B1%20%5C%20mol%20%5C%20FeCl_3%7D%7B162.19%20%5C%20g%20%5C%20FeCl_3%7D)
![345.0 *\frac{1 \ mol \ FeCl_3}{162.19 }](https://tex.z-dn.net/?f=345.0%20%2A%5Cfrac%7B1%20%5C%20mol%20%5C%20FeCl_3%7D%7B162.19%20%7D)
![\frac{345.0 \ mol \ FeCl_3}{162.19 }](https://tex.z-dn.net/?f=%5Cfrac%7B345.0%20%5C%20mol%20%5C%20FeCl_3%7D%7B162.19%20%7D)
Divide.
![2.12713484 \ mol \ FeCl_3](https://tex.z-dn.net/?f=2.12713484%20%5C%20mol%20%5C%20FeCl_3)
3. Round
The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.
For the answer we calculated, that is the thousandth place.
The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.
![\approx 2.127 \ mol \ FeCl_3](https://tex.z-dn.net/?f=%5Capprox%202.127%20%5C%20mol%20%5C%20FeCl_3)
There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.
It evaporates so the ice cube turns in to water then after words evaporat
Answer:
a) Step 1:
![2PbS(g)+3O_2(g)\overset{roasting}\rightarrow 2PbO(s)+2SO_2(g)](https://tex.z-dn.net/?f=2PbS%28g%29%2B3O_2%28g%29%5Coverset%7Broasting%7D%5Crightarrow%202PbO%28s%29%2B2SO_2%28g%29)
Step 2:
![2PbO(s)+PbS(s)\overset{\Delta }\rightarrow 3Pb(l)+SO_2(g)](https://tex.z-dn.net/?f=2PbO%28s%29%2BPbS%28s%29%5Coverset%7B%5CDelta%20%7D%5Crightarrow%203Pb%28l%29%2BSO_2%28g%29)
b) The overall balanced reaction for given process is ;
![3PbS(s)+3O_2(g)\rightarrow 3Pb(l)+3SO_2(g)](https://tex.z-dn.net/?f=3PbS%28s%29%2B3O_2%28g%29%5Crightarrow%203Pb%28l%29%2B3SO_2%28g%29)
Explanation:
a)
Galena = ![PbS](https://tex.z-dn.net/?f=PbS)
Lead(II) oxide = ![PbO](https://tex.z-dn.net/?f=PbO)
Sulfur dioxide = ![SO_2](https://tex.z-dn.net/?f=SO_2)
Step 1:
Roasting the galena in oxygen gas to form lead(II) oxide and sulfur dioxide.
Balanced equation of step 1:
..[1]
Step 2:
Heating the metal oxide with more galena forms the molten metal and more sulfur dioxide.
Balanced equation of step 2:
..[2]
b)
For over all reaction add [1] and [2]. The overall balanced reaction for given process is ;
![3PbS(s)+3O_2(g)\rightarrow 3Pb(l)+3SO_2(g)](https://tex.z-dn.net/?f=3PbS%28s%29%2B3O_2%28g%29%5Crightarrow%203Pb%28l%29%2B3SO_2%28g%29)