n=RT/PV finding a mass? [chem]?

Question

Hi! So I am doing a prelab for my lab today and I am having problems with figuring out if my answer is correct or not.

Calcium carbonate reacts with hydrochloric acid as follows:

CaCO3(s) + 2HCL(aq) = CaCl2(aq) +H2O(l) + CO2 (g)

What mass of CaCO3 must react with excess HCl to produce 10.0L of CO2 gas at 25 degree C and 74.0 cm Hg pressure?

B) Calculate the density of chlorine gas at 2.55 atm pressure and a temperature of 300 degree C.

Answer 1

*** a ***

PV = nRT

n = PV/(RT)



so..

n CO2 = (74.0cmHg x 1atm/76.0cmHg) x (10.0L) / ((0.08206 Latm/molK) x (298.15K))

n CO2 = __ mol... ?



then..

__ mol CO2 x (1 mol CaCO3 / 1 mol CO2) x (__g CaCO3 / mol CaCO2) = __g CaCO3



*** b ***

PV = nRT

n = mass / mw



sub and rearrange

PV = (mass / mw) RT

(mass / V) = mw x P / (RT)



since density = mass / V

density = mw x P / (RT)



solve..

density = (70.9 g/mol) x (2.55atm) / ((0.08206 Latm/molK) x (573.15K)) = __ g/L



********

I'll leave the rest for you

Answer 2

A)

Let P = the pressure = 74.0 cm Hg = 740 mm Hg

Let V = the volume = 10 L

Let T = the temperature = 25 C = 298 K

Let R = the gas constant = 62.36367 (L•mm Hg)/(K• mole)

Let n = the number of moles of CO2



Using the gas law, (P)(V) = (n)(R)(T), solved for n:

n = {(P)(V)}/{(R)(T)}

n = {(740 mm Hg)(10 L)}/{(298 K)(62.36367 L•mm Hg/K• mole)}

n ≈ 0.3981 moles

Because Calcium Carbonate reacts to produce CO2 in a 1:1 ratio, 0.3981 moles of Calcium Carbonate are needed to produce 0.3981 moles of CO2. The molar mass of Calcium Carbonate is 100 grams per mole (google search)



0.3981 moles(100 grams/mole) = 39.81 grams of Calcium Carbonate.



B)

Density is mass/volume. To make the calculation simple, assume that you have 1 mole and then find its volume. Chlorine gas is diatomic so its molar mass is twice its atomic weight = 70.9 grams/mole. The temperature must be converted to 573 K and the we use the gas law to find the volume but this time we use 0.082 {L•atm}/{K•mole} for the gas constant:



V = n(R)(T)/P

V = 1(0.082)(573)/2.55

V ≈ 18.43 L



D = 70.9 grams/18.43 L



D ≈ 3.9 grams/Liter

Answer 3

ok. by using fact that this could properly be a decomposition there is not any could prefer to locate the limiting reagent. The 4.80 2 grams is beside the point till at last the tip. So first you have 10 grams of C6H1206, which has a molar mass of one hundred 80. all of us be responsive to that C2H5OH molar mass is 40 six. one way of doing it truly is putting 10g/180g = x/ninety 2 (this could properly be a ratio that takes below attention the completed quantity of mass waiting to be made out of 10g of glucose the situation x is the completed grams of ethanol produced. This ninety 2 comes from multiplying 40 six x 2 by using fact we've 2 moles of ethanol) x is chanced directly to be 5.1111g. Then the theoretical yield could be 4.80 2/5.111=.943