Using a sample of 0.385 g of Alka-Seltzer, the mass of CO 2 is found to be 0.122

Question

Using a sample of 0.385 g of Alka-Seltzer, the mass of CO2 is found to be 0.122 g.

1. Calculate the mass of NaHCO3 that produced the 0.122 g of CO2

2. The % mass of NaHCO3 in the sample

3.The mass of NaHCO3 in the tablet assuming the mass of the tablet is 3.50 g

If the data is

Mass (before reaction): test tube + HCl(aq) + stir bar + capsule = 26.042 g
Mass (after reaction): test tube + HCl(aq) + stir bar + capsule = 25.783 g
Volume of water displaced from the squirt bottle = 144 mL
Temperature of the CO2(g) = 293.6 K
CO2(g) pressure = 0.986 atm

Calculate the mass of the CO2(g)

Calculate the moles of CO2(g)

Calculate the density of CO2 in g/lL

Calculate the molar mass of CO2 assuming a temperature of 293.6 K

Explanation / Answer

NaHCO3 + HCl ---------- CO2 + NaCl + H2O

0,122g CO2/ 44g/mol= 0,003mol CO2

1 mol NaHCO3 -----produce----- 1mol CO2

X mol ----produce----- 0,003mol CO2

X= 0,003mol NaHCO3* 84g/mol =0,252gNaHCO3

0,385g mass of the sample -------------- 100%

0,252 g NaHCO3 -------------- X = 65,45%

If the tablet is 3,50g ----------- 100%

X gNaHCO3 ----------- 65,45%

X= 2,29g NaHCO3

-----------------------------------------------------------------------------------------------------------------------------------------------------------

a) before - after reaction = 0,259g give us the mass of CO2 that produces in the reaction because the CO2 is a gas we loss the mass of the solution because scapes from it.

b) 0,259g / 44g/mol = 0,0058mol CO2

c) density = mass/ volumen (L) = 0,259g/0,144L = 1,80 g/L

d) PV=nRT ------------ n/V= P/RT= 0,986atm/ 0,082Latm/Kmol * 293.6K = 0,041mol/L= 0,041M

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.