Answer:
See explanation.
Explanation:
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In this case, according to the description of the chemical reaction, it is possible to write the corresponding equation as follows:
[tex]Mg+2HCl\rightarrow MgCl_2+H_2[/tex]
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:
[tex]m_{H_2}^{by\ Mg}=23gMg*\frac{1molMg}{24.305gMg}*\frac{1molH_2}{1molMg}=0.95gH_2 \\\\m_{H_2}^{by\ HCl}=0.412L*2.6\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}*=0.54gH_2[/tex]
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:
[tex]V_{H_2}=\frac{0.54molH_2*0.08206\frac{atm*L}{mol*K}*(81+273.15)K}{1.7625atm} \\\\V_{H_2}=8.9L[/tex]
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:
[tex]n_w=\frac{0.4675atm*8.9L}{0.08206\frac{atm*L}{mol*K}*(81+273.15)K} \\\\n_w=0.143molH_2O\\\\m_w=0.143molH_2O*\frac{18.02gH_2O}{1molH_2O} =2.6gH_2O[/tex]
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