Question

1. Butane, C4H10, reacts with oxygen, O2, to form water, H2O, and carbon dioxide, CO2, as shown in the following chemical equation:

2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g)

Calculate the mass of butane needed to produce 71.5 g of carbon dioxide.

2. For the reaction, calculate how many grams of the product form when 22.4 g of Ca completely reacts.
Assume that there is more than enough of the other reactant.
Ca(s)+Cl2(g)→CaCl2(s)

3. For the reaction, calculate how many grams of the product form when 22.4 g of Br2 completely reacts.
Assume that there is more than enough of the other reactant.
2K(s)+Br2(l)→2KBr(s)

4. For the reaction, calculate how many grams of the product form when 22.4 g of O2 completely reacts.
Assume that there is more than enough of the other reactant.
4Cr(s)+3O2(g)→2Cr2O3(s)

5. For the reaction, calculate how many grams of the product form when 22.4 g of Sr completely reacts.
Assume that there is more than enough of the other reactant.
2Sr(s)+O2(g)→2SrO(s)

6. In Part A, you found the amount of product (3.40 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Part B, you found the amount of product (3.00 mol P2O5 ) formed from the given amount of oxygen and excess phosphorus.

Now, determine how many moles of P2O5 are produced from the given amounts of phosphorus and oxygen.

Express your answer to three significant figures and include the appropriate units. P4+5O2→2P2O5

7. How much heat is produced by the complete combustion of 205 g of CH4?
CH4(g)+2O2(g)→CO2(g)+2H2O(g)ΔH∘rxn=−802.3kJ

Answer 1

the reaction is

2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g), the stoichiometry suggests that 2mole of butane( limiting reactant) reacts with excess oxygen (the excess reactant) to give 8 moles of CO2

mole =mass/molar mass

molar mass data ( g/mole): C4H10=58, O2= 32, H2O= 18 and CO2=44

8*44=352 gm of CO2 is produced from 2*58=116 gm of C4H10

71.5 gm of CO2 is produced from 116*71.5/352 =23.56 gm of C4H10

2. The reaction is Ca(s)+Cl2(g)→CaCl2(s), atomic weights : Ca= 40, Cl2= 71 and CaCl2=111

the stoichiometry of the reaction suggests that 40 gm of Ca completely reacts to give 111 gm of CaCl2, product.

22.4 gm of Ca completely reacts to give 111*22.4/40 gm of CaCl2=62.14 gm of CaCl2.

3. For the reaction, calculate how many grams of the product form when 22.4 g of Br2 completely reacts.
Assume that there is more than enough of the other reactant.

atomic weights : K= 39, molar mass of Br2= 160

The reaction is 2K(s)+Br2(l)→2KBr(s)

2 mole of K reacts with 1 mole of Br2 to give 2 mole of KBr.

2*39=78 gm of K reacts with 160 gm of Br2 to give 2*(39+80)= 238 gm of KBr

22.4 gm of Br2 reacts to give 22.4*238/160 gm of KBr=33.32 gm of KBr.

4. Atomic weights : Cr= 52 and O2= 32

4Cr(s)+3O2(g)→2Cr2O3(s) reaction suggests that 4*52=208 gm of Cr reacts with 3*32= 96 gm of O2 to 2*152=304 gm of Cr2O3.

22.4 gm of O2 gives 304*22.4/96 =70.93 gm of Cr2O3.

5.atomic weights : Sr= 88 gm/mole, O2= 32

2Sr(s)+O2(g)→2SrO(s) suggests 2 moles = 2*88=176 gm of Sr reacts completely to give 2*(88+16)=208 gm of SrO.

22.4 gm of Sr completely reacts to give 22.4*208/176 =26.47 gm of SrO.

6. molecular mass : P4= 31*4=124 , O2= 32

the reaction is P4+5O2→2P2O5

let x= mole of P4, since oxygen is excess and moles of P4 is limited, mole of P2O5 formed =2* moles of P4=2x=3.4

x= 1.7 moles

in the seconc case, let y= moles of oxygen, the lmiting reactant.

5 moles of O2 reacts with P4 to give 2 moles of P2O5.

moles of O2 required for forming 3 moles of P2O5= 5*3/2= 7.5 moles, hence y=7.5

moles of P4= 1.7 and y= 7.5

molar ratio of P4: O2 ( theoretical)= 1:5, actual ratio = 1.7:7.5 = 1:4.4

hence oxygen is limiting reactant in this scenariio and 5 moles of O2 gives 2 mole of P2O5

7.5 moles of O2 gives       2*7.5/5= 3 moles of P2O5.

molar mass of P2O5= 284 g/mole, mass of P2O5= moles* molar mass = 3*284= 852 gm

7. from the reaction, CH4(g)+2O2(g)→CO2(g)+2H2O(g)ΔH∘rxn=−802.3kJ

1 mole of CH4 gives 802.3 Kj of heat

molar mass of CH4= 16 g/mole

1 mole of CH4= 16 gm of CH4

16 gm of CH4 gives 802.3 Kj

205 gm of CH4 gives 205*802.3/16= 10279.47 KJ heat is produced.

Now, determine how many moles of P2O5 are produced from the given amounts of phosphorus and oxygen.

Express your answer to three significant figures and include the appropriate units. P4+5O2→2P2O5