Kindly assist with step by step solution. Thank you (a) A mass of 0.4113 g of an

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Kindly assist with step by step solution.

Thank you

(a) A mass of 0.4113 g of an unknown acid, HA, is titrated with NaOH(aq). If the acid reacts with 28.10 mL of 0.1055 M NaOH(aq), what is the molar mass of the acid? 1. (b) How many grams of KCIOs are needed to produce of 2.56 grams of O2? 2KCIO3(s) 2KCI(s)+302(g) (c) A bicycle tire is filled to a pressure of 4.42 atm (65 psi) at a temperature of 12degree C. If the temperature of the tire increases to 33 degree C, what is the pressure in the tire? Assume the volume of the tire is constant. (d) If 8.5 g of oxygen gas (O2) is introduced into an evacuated 1.50 L flask at 22°C, what is the pressure inside the flask? (e) At 338 mm Hg and 72°C, a sample of carbon monoxide gas occupies a volume of 0.225 L. The gas is transferred to a 1.50-L flask and the temperature is reduced to -15°C. What is the pressure of the gas in the flask?

Explanation / Answer

The reaction between acid (HA) and NaOH is

HA+ NaOH----à NaA + H2O

1. 1 mole of HA requires 1 mole of NaOH as per the reaction.

Moles of NaOH= molarity* volume in L= 0.1055*28.10/1000

This is also the moles of HA

Hence moles of HA =0.1055*28.10/1000

Moles of HA= mass/molar mass =0.4133/M , M= molar mass

0.4133/M= 0.1055*28.10/1000

M= 0.4133*1000/(0.1055*28.1)=139 g/mole

2 The reaction of decompositon of KClO3 is 2KClO3-------->2KCl + 3O2

2 moles of KClO3 produces 3 moles of Oxygen.

atomic weights : K= 39, Cl =35.5, O= 16

molar mass of KClO3 (g/mole)= 39+35.5+3*16= 122.5 , molar mass of oxygen =2*16=32

as per the stocihiometry of th reaction, 3 moles of oxygen requires 2 moles of KClO3

or 3*32= 96 gm of oxygen is produced from 2*122.5= 245 gm of KClO3

2.56 gm of oxygen is produced from 245*2.56/96 =6.53 gm of KClO3

3. Gas law equation can be written as PV/T= constant ( when moles of gas is constant)

at constanr volume . P/T= constant

or P1/T1= P2/T2, where P1= 4.42 atm, T1= 12 deg.c whicn need to be converted to K, T= 12+273= 285

T2= 33deg.c= 33+273= 306K, P2= P1T2/T1= 4.42* 306/285 atm =4.745 atm

4. Gas law equation can be writen as PV=nRT

P= pressrue in atm, V= 1.5 L, n= moles of Oxygen = mass/molar mass =8.5/32= 0.2656, R= 0.0821 L.atm/mole.K

T= 22deg.c= 22+273 =295K

P= nRT/V=0.2656*0.0821*295/1.5 =4.288 atm

5. Given P= 338 mm Hg, P in atm =338/760   atm = 0.44 atm, T=72 deg.c= 72+273= 345K, V= 0.225 L

n= PV/RT= 0.44*1.5/(0.0821*345)= 0.0233

when transferred to 1.5 L container, no of moles remains constant

when T= -15 deg.c= -15+273= 258K, n=0.0233, V= 1.5L, P= nRT/V= 0.0233*0.0821*258/1.5 atm =0.33 atm =0.33*760 mm Hg= 250.8 mm Hg

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