Question

If you started with 23g of Mg and 412 ml of 2.6M HCl, find the limiting reactant. If you collected the hydrogen gas over water at a temperature of 81OC, what is the volume of just the water if the total pressure is 2.23 atm? How many grams of water vapor are in the container?

Answer 1

See explanation.

Step-by-step explanation:

Hello there!

In this case, according to the description of the chemical reaction, it is possible to write the corresponding equation as follows:

`Mg+2HCl\rightarrow MgCl_2+H_2`

Whereas we are given the mass of magnesium and the concentration and volume of HCl; it means that we can calculate the moles of hydrogen yielded by each of these reactants in order to identify the limiting reactant:

`m_(H_2)^(by\ Mg)=23gMg*(1molMg)/(24.305gMg)*(1molH_2)/(1molMg)=0.95gH_2 \\\\m_(H_2)^(by\ HCl)=0.412L*2.6(molHCl)/(1L)*(1molH_2)/(2molHCl)*=0.54gH_2`

Therefore, we infer that the limiting reactant is HCl as it yields the fewest moles of hydrogen. Next since the vapor pressure of water at 81 °C is about 0.4675 atm we infer that the pressure of hydrogen is 2.23 atm - 0.4675 atm = 1.7625 atm. In such a way, we use the ideal gas equation to obtain the volume of hydrogen:

`V_(H_2)=(0.54molH_2*0.08206(atm*L)/(mol*K)*(81+273.15)K)/(1.7625atm) \\\\V_(H_2)=8.9L`

Finally, since the pressure of water is 0.4675 atm and the volume is the same for both, we obtain the moles of water and subsequently the required grams as shown below:

`n_w=(0.4675atm*8.9L)/(0.08206(atm*L)/(mol*K)*(81+273.15)K) \\\\n_w=0.143molH_2O\\\\m_w=0.143molH_2O*(18.02gH_2O)/(1molH_2O) =2.6gH_2O`

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