Show the calculation for the mass of magnesium metal requiredto produce 40.0 ml

Question

Show the calculation for the mass of magnesium metal requiredto produce 40.0 ml of h2 with a temp of 27.6 C and an atm pressureof 749 mmHG. use equation 1 and 3. Equation 1 - Mg(s) + 2HCL (aq) --> MgCl2 (aq) + H2(g) equation 3 - nh2 = (Patm - Ph20) Vb/RT Show the calculation for the mass of magnesium metal requiredto produce 40.0 ml of h2 with a temp of 27.6 C and an atm pressureof 749 mmHG. use equation 1 and 3. Equation 1 - Mg(s) + 2HCL (aq) --> MgCl2 (aq) + H2(g) equation 3 - nh2 = (Patm - Ph20) Vb/RT

Explanation / Answer

First make sure the equation is balance, which is looks likeit is Mg(s) + 2HCL (aq) --> MgCl2 (aq) + H2 (g) now convert 749 mmHg to atm, ml to liters and 27.6 tokelvin 749 mmHg/760 torr = 0.99 atm 40ml/1000 = 0.040 L 273.15 K+ 27.6 = 300.75K now you have all the correct units. Now you just plugeverything into Pv=nRT (R is the ideal constant at 0.08206) (0.99)(0.040) = (n)(0.08206)(300.75) n = 0.997 moles now that we have mole, we can find mass of Mg by usingdimensional analysis and the molar mass of Mg (0.997 moles)(1 mol Mg/ 2 mol H2)(24.31 molar mass of Mg) =11.8 g of Mg hope this helped. Don't forget to give karma please

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