2Al (s) + 3MnSO4 (aq) --> Al2(SO4)3 (aq) + 3 Mn (s) 1.) a 5.00 gram sample of al

Question

2Al (s) + 3MnSO4 (aq) --> Al2(SO4)3 (aq) + 3 Mn (s)

1.) a 5.00 gram sample of aluminum powder is known to be contaminated with some organic solid. An analysis of the sample required 47.50 mL of a 0.250 M MnSO4 solution. What is the percentage of Aluminum in the sample?

2.) A chunk of dry ice (solid carbon dioxide) is allowed to sublime ( convert from solid to gas) into a large balloon. What is the mass of the dry ice if the final volume of the balloon is 17.5 L at a pressure and temperature of 0.980 atm and 23 degrees C respectively?

(EXTRA Question for max points)

A sample of a gas occupies 755mL at 25 degrees C and a pressure of 185 mmHg. If the temperature is raised to 70.0 degrees C and the pressure is increased to 330 mmHg, what is the new volume of the gas?

760mmHg = 1 atm

Show all work please and thank you!

Explanation / Answer

a) moles of aluminum present = 47.5*0.25*10^-3*(2/3)

=7.916*10^-3

so mass of aluminum present = 7.916*10^-3*27

=0.213 g

so mass % of Al = 0.213*100/5

=4.26%

2)using PV=nRT,

0.98*17.5= (m/44)*0.0821*(273+23)

or m=31.05 grams

3)let the volume be V ml. so,

using PV=nRT

or 755*185/(273+25) = v*330/(273+70)

or v=487.172 mL

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