Can you help me with this problem about a magnesium metal reaction?

Question

Magnesium metal is reacted with excess HI to produce hydrogen gas. The gas is collected over water. 45.82mL of the wet hydrogen gas is collected at 30.0°C. The water level in the eudiometer is equalized with water in another vessel before the volume of H2 gas is determined. The atmospheric pressure in the laborotory is 0.972 atm.

 

A. What is the vapor pressure of water under these conditions?

 

B. What is the partial pressure of the H2 gas

 

C. What mass of Mg metal was consumed, assuming 100% yield

Answer 1

In order to determine the answers to B & C, you must have the answer to A.  From the given data, you cannot "calculate" this value; it should be available to you via a table of water vapor pressures at various temperatures.  According to a table that I have, the vapor pressure of H2O at 30.0 ^oC is 31.82 torr.

 

Now,Using Dalton's Law of Partial Pressures, you can calculate the partial pressure of the "dry" H2 gas.

Ptotal = P H2 + P H2O

I'll convert the given atm to torr 1st.  (760 torr/atm)(0.972 atm) = 738.7 torr

So, 738.7 torr =  P H2 + 31.82 torr; P H2 = 706.9 torr

 

Now, for part C.  The balanced equation is:  Mg + 2HI  --> MgI2  +  H2, so the molar ratio of Mg to H2 is 1:1

You can use the ideal gas law to determine the moles of H2 gas first, then from the stoichiometry find the mass of Mg consumed.

 

PV = nRT so, n = PV/RT = [(706.9 torr)(1 atm/760 torr)][0.04582 L]/[(0.0821 atm L/mol K)(303 K)] = 1.71 x 10^-3 mol

 

Since this is the same as the moles of Mg, we can now calculate the mass of the Mg:

(24.3 g Mg/mol) (1.71 x 10^-3 mol) = 4.16 x 10^-2 g