#1. For any given object, can mass and/or weight vary? Explain.
#2. How would you calculate the molecular weight of a compound like `Fe(C_5H_5)_2`?
#3. How many molecules of C are in the molecule `C_4H_8`?
#4. In any given reaction, if the mass of reactants is 80g, what must the mass of the products be?
#5. In simplest terms, what does the law of conservation of mass say?
#6. Conceptually, what is a limiting reagent?
#1. Balance the following chemical equation:
`P_4+O_2 rArr P_2O_5`
#2. Balance the following chemical equation:
`AlBr_3+Cl_2 rArr AlCl_3+Br_2`
#3. Balance the following chemical equation:
`C_5H_8O_2+NaH+HCl rArr C_5H_(12)O_2+NaCl`
#4. Convert 5 moles B to number of atoms.
#5. Convert 75g of `ClF_3` to number of atoms.
#6. Convert 500g of GaAs to moles.
#7. You're given 150g of both `C_4H_10O` and `O_2` for the following reaction:
#8. A molecule with the formula `C_xH_yO_z` is composed of 60.0% C, 13.4% H, and 26.6% O. Determine the formula of the molecule by solving for x, y, and z.
#9. For the following equation, you're given 145g `C_4H_(10)` and 65g `O_2`:
`C_4H_10+ O_2 rArr CO_2 +H_2O`.
a) Balance the reaction.
b) Which reactant is the limiting reactant? Which reactant is in excess?
c) How many grams of `CO_2` are produced? How many grams of `H_2O`?
d) How many grams of the excess reactant remain?
e) Check your answer using the law of conservation of mass.
#1. For any given object, weight can vary, but mass must remain the same. This is because mass is an inherent property of an object whereas weight can vary depending on gravity.
#2. To calculate the molecular weight of a compound, simply add together the molecular weight of each individual atom. In this case, there is 1 `Fe`, 10 `C`, and 10 `H`. Therefore, `MW_(Fe(C_5H_5)_2)=m_(Fe)+10m_C+10m_h= 55.845 "amu"+10(12.011 "amu")+10(1.008 "amu")=186 "amu"`
#3. 4. The subscript next to an element denotes the number of atoms of that element.
#4. If the mass of the reactants is `80 "g"`, the mass of the products must be `80 "g"` as well. This is because mass is conserved in chemical reactions.
#5. Mass is neither created nor destroyed.
#6. A limiting reagent in a chemical reaction is the reactant that runs out first.
#1. `P_4+5O_2 rArr 2P_2O_5`
#2. `2AlBr_3+3Cl_2 rArr 2AlCl_3+3Br_2`
#3. `C_5H_8O_2+2NaH+2HCl rArr C_5H_(12)O_2+2NaCl`
#4. `(5 "mol" B)((6.022*10^(23) "atoms")/(1 "mol" B))=3.01*10^(24) "atoms"`
Note that we didn't have to use the molecular weight of `B`. This is because 5 moles of anything will be equal to 5 moles of anything else, since a mole is just a number; The number of objects in a dozen donuts is the same as in a dozen cars. Same principle.
#5. `(75 g ClF_3)((1 "mol" ClF_3)/(92.448 g ClF_3))((6.022*10^(23) "atoms")/(1 "mol" ClF_3))=4.89*10^(23) "atoms" ClF_3`
#6. `(500 g GaAs)((1 "mol" GaAs)/(166.4 g GaAs))=3 "mol" GaAs`
#7. `(350 g SiO_2)((1 "mol" SiO_2)/(60.08 g SiO_2))((1 "mol" Si)/(1 "mol" SiO_2))=5.83 "mol" Si`
#8. Assume that you're starting off with `100"g"` of the compound. That means you have `60"g" C`, `13.4"g" H`, and `26.6"g" O`. Convert these into moles and you'll get `5 "moles" C`, `13 "moles" H`, and `1.66 "moles" O`. Since you have `C_5H_(13)O_(1.66)`, divide all of the values by 1.66 to get whole numbers: `C_(5/1.66)H_(13/1.66)O_(1.66/1.66)= C_3H_8O`.
#9. a) `2C_4H_10+13O_2 rArr 8CO_2+10H_2O`
b) First convert to moles
`(145 g C_4H_(10))((1 "mol")/(58.1 g C_4H_(10)))=2.5 "mol" C_4H_(10)`
`(64 g O_2)((1 "mol")/(32 g O_2))=2 "mol" O_2`
Determine the conversion factors of reactants.
`(2 "mol" O_2)((2 "mol" C_4H_(10))/(13 "mol" O_2))=0.308 "mol" C_4H_(10)`
`(2.5 "mol" C_4H_(10))((13 "mol" O_2)/(2 "mol" C_4H_(10)))=16.25 "mol" O_2`
`O_2` is the limiting reactant because the reaction requires more `O_2` than is present in the reaction.
c) Convert the moles `O_2` to moles final product using the conversion factors.
`(2 "mol" O_2)((8 "mol" CO_2)/(13 "mol" O_2))=1.23 "mol" CO_2`
`(1.23 "mol" CO_2)((44.01 g CO_2)/(1 "mol" CO_2))=54.1 g CO_2`
`(2 "mol" O_2)((10 "mol" H_2O)/(13 "mol" O_2))=1.54 "mol" H_2O`
`(1.54 "mol" H_2O)((18.016 g H_2O)/(1 "mol" H_2O))=27.7 g H_2O`
d) To find out how many grams of the excess remain, convert the amount of reacted `O_2`, first convert moles of `O_2` to moles `C_4H_(10)`
`(2 "mol" O_2)((2 "mol" C_4H_(10))/(13 "mol" O_2))=0.31 "mol" C_4H_(10)`
Now, convert that amount of moles to grams `C_4H_(10)`
`(0.31 "mol" C_4H_(10))((58.1 g C_4H_(10))/(1 "mol" C_4H_(10)))=17.9 g C_4H_(10)`
Subtract this amount used from the initial amount of `C_4H_(10)`
`145 g - 17.9 g=127.1 g` left over.
e) To check whether our answer is correct or now, we can apply the law of conservation of mass. The mass of reactants should be equal to the mass of the products.
Mass reactants: `m_(O_2)=(2 "mol" O_2)((32 g O_2)/(1 "mol" O_2))=64 g O_2`
`m_(O_2)+m_(C_4H_(10))=64g+17.4g=81.4 g`
Mass products: `m_(CO_2)+m_(H_2O)=54.1g+26.7g=80.8 g`
Since I rounded a lot of the numbers, this answer makes sense. Accounting for the rounding, the mass of the reactants is the same as the mass of the products. Notice that this is not the total mass of the reaction, as `127.1 "g"` of the `C_4H_(10)` was unreacted. That amount does not count in the calculation of conservation of mass, seeing as it just sits there.