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I need some help on the following question.


Calcium Carbonate reacts with hydrochloric acid to release carbon dioxide according to the following equation.

CaCO^3 + 2HCl = CO^2 + H^2O + CaCl^2

Atomic weights Ca = 40.1 , C = 12.0, O = 16.0 , Cl = 35.5 , H = 1.0

a. If 12.5 grams of calcium carbonate are mixed with an excess of acid calculate the moles of CO^2 formed.

b. Calculate the volume of CO^2 ( at STP) produced from the 12.5 grams of Clacium Carbonate.

c.If the 12.5 grams of CaCO^3 had been mixed with a volume of HCl such that the number of moles of HCl was 0.2 moles, calculate the moles of CO^2 formed



Any contribution will be appreciated.

Thankyou

Answer 1

A.

(1) calculate moles CaCO3 (fyi, it isn't CaCO^3 = CaCO³. the 3 is a subscript. the ^ usually means superscript. most of us with a chemistry background would just write CaCO3 on a computer instead of trying to deal with making subscripts....)

(2) convert moles CaCO3 to moles CO2. note that your balanced equation compares the number of molecules of CaO3 to CaO2 which is also the number of moles of each. 1 mole of CaCO3 ----> 1 mole CO2....

so here we go....

moles = weight / mw.
weight = 12.5 g
mw = atomic mass Ca + atomic mass C + 3 x atomic mass O = 40.1+12.0+3x16.0 = 100.1 g/mole

so moles CaCO3 = 12.5g / (100.1 g/mole) = 0.125 moles

then moles CO2 = 0.125 moles CaCO3 x (1 mole CO2 / 1 mole CaCO3) = 0.125 moles CO2.

B.

assume ideal gas. Use PV = nRT. Calculate V.....
also STP means either T = 273.15K and P = 100 kPa or it means T = 273.15k and P = 101.325 kPa. I'll assume the first. you can change it if you need to....

V = nRT/P

n=0.125 moles
T = 273.15k
P = 100 kPa
R= 8.314 L kPa/(mole K)

V = 0.125 moles x 8.314 L kPa/(mole K) x 273.15K / 100 kPa
= 2.84 L

C.

Ok.. you have 0.125 moles CaCO3. since you need 2 moles HCl per 1 mole CaCO3, you need 0.250 moles HCl to react with all the CaCO3. since you have less than that, HCl is the "limiting reagent" and that needs to be the basis of your calculations.....

0.20 moles HCl x (1 mole CO2 / 2 moles HCl) = 0.10 moles CO2.

note that the units of "moles HCl" is on the top and bottom of the equation and therefore cancel. leaves moles CO2....

and 1 mole CO2 = 2 moles HCl from balance equation. dividing both sides by 2 moles HCl gives....

1 mole CO2 = 2 moles HCl
1 mole CO2 / 2 moles HCl = 2 moles HCl / 2 moles HCl

1 mole CO2 / 2 moles HCl = 1

and since any number x 1 = that number then......

0.20 moles HCl x 1 = 0.20 moles HCl x (1 mole CO2 / 2 moles HCl) and presto you get moles CO2.

process is called unit factor method. google it. very useful.

Answer 2

a. Use the 12.5 g to calculate the # of moles of CaCO3. Then use the mole ratio (from the balanced eqn) to calculate moles of CO2 (1:1)

12.5 g CaCO3 ( 1 mol/100.1 g) = 0.125 mol CaCO3

0.125 mol CaCO3 (1 mol CO2/1 mol CaCO3)=0.125 mol CO2

b) At STP, we know that there are 22.4 L per mole of gas (any gas). Since you know that 12.5 g CaCO3 makes 0.125 mol of CO2:

0.125 mol CO2 (22.4 L/mol) = 2.80 L of CO2

c) now you have a limiting reactant problem...if there is not sufficient HCl to completely react with the CaCO3, then the reaction will stop when the HCl is completely used up. So, determine the moles of CO2 from it. If it is less than the 0.125 mol CO2 calculated in (a), then you must determine the volume from it.

0.200 mol HCl (1 mol CO2 / 2 mol HCl) = 0.100 mol CO2

This is a reduction in the # of moles (HCl is limiting reactant):

0.100 mol CO2 ( 22.4 L / mol CO2) = 2.24 L CO2

Answer 3

a. CaCO3 + 2H+ ===> Ca++ + CO2 + H2O

Formula weight CaCO3=100

12.5gCaCO3 x 1molCaCO3/100gCaCO3 x 1molCO2/1molCaCO3 = 0.125 moles CO2

b. 0.125molCO2 x 22.4LCO2/1molCO2 = 2.8L CO2

c. 12.5gCaCO3 x 1molCaCO3/100gCaCO3 = 0.125 moles CaCO3

The same amount of CO2 would form, because 0.2 mole HCl is more than the 0.125 mole needed to react with the amount of CaCO3. The CaCO3 is what is called the limiting reagent: It will run out first.