1) Write the complete balanced, molecular equation for the reaction that produce
ID: 829470 • Letter: 1
Question
1) Write the complete balanced, molecular equation for the reaction that produces the carbon dioxide.
2) How many moles of CaC03 are in the 30.0 grams of CaCO3 used?
3) How many liters of CO2 can be generated by this mass of CaCO3 at STP?
4) What is the mass of one Liter of each of the following gases at STP.
*Use only the Atomic Masses and the Molar Volume*
a) Fluorine
b) Ammonia
c) Propane(c3H8)
5) A 1.60 gram sample of impure sodium acetate was mixed with excess sodium hydroxide and heat producing solid sodium carbonate and methane gas.
a) Write the complete, balanced, molecular equation for the reaction.
b) The methane was collected over water at 27 C and 747 Torr, producing a total volume of 320.ml of as. Calculate the number of moles of methane fas that was collected. *You will need to correct for the vapor pressure of the water
c) Calculate the mass of sodium acetate that would be required to produce this amoung of methane
d) Calculate the percent of pure sodium acetate in the impure sample
Explanation / Answer
1) NaHCO3 + HC2H3O2 ---> NaC2H3O2 + H2O + CO2
2) 40 + 12 + 48 = 100 (Mr of CaCO3) hence no. of moles = 30/100 = 0.3 moles
3) Use dimension analysis to solve the problem. The molar mass of CaCO3 is 100g and there are 22.4 Liters in a mole at STP
30 g CaCO3 ( 1 mole CaCO3/ 100g CaCO3) (1 mole of CO2/ 1 mole CaCO3) (22.4 L/ 1 mole CO2)
= 0.3 * 22.4 = 6.72 L CO2
4) a) A mole of gas at STP occupies 22.4 L.
So 1 L of F2 is 0.0446 moles of F2. (1/22.4)
To find grams just mulitply by the molar mass:
0.0446 moles * (19*2 g/mol) = 1.69 g
b)22.4 litres of any gas at S.T.P. weigh equal to the molecular mass of gas expressed in grams...this is called Gram-molecular volume law..
so 22.4 litres of NH3 at STP have mass = 17g
so 1 litre of NH3 will have mass = 1722.4 = 0.758 g
c) 22.4 litres of any gas at S.T.P. weigh equal to the molecular mass of gas expressed in grams...this is called Gram-molecular volume law..
so 22.4 litres of propane at STP have mass = 44g (molecular mass of propane = 12 X3 + 8 X 1 = 44 carbon's atomic mass is 12 and hydrogen's atomic mass is 1..)
so 1 litre of propane will have mass = 44/22.4 = 1.96 g
5) a) C2H3NaO2 + NaOH --> Na2CO3 + CH4
b) Use PV=nRT, the ideal gas law. P is pressure (atmospheres), V is volume in liters, n is moles, R is Faraday's Constant (0.08206) and T is temp in Kelvin (C+273). Plug in the numbers and solve for x.
747 torr = 0.983 atmospheres (747/760)
(0.983) (0.32) = n (0.08206) (300)
0.315 = 24.62n
n = 0.013 moles
c) MM of sodium acetate = 82 now moles of methane = moles of sodium acetate hence mass of sodium acetate required = 0.013*82 = 1.066 gms
d) percentage of pure sodium acetate = (1.066 * 100)/1.6 = 66.7 percent
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