How many grams of Mg is needed to create 762.44 mL of H2 gas at 39.86oC and  614 mmHg according to the reaction below.Mg + 2HCl --> MgCl2 + H2(g)

To solve this problem, we need to use the ideal gas law and stoichiometry.

Step 1: Convert the given conditions to the appropriate units. Temperature must be in Kelvin and pressure in atm for the ideal gas law.

T = 39.86°C + 273.15 = 313.01 K P = 614 mmHg * (1 atm / 760 mmHg) = 0.80789 atm

Step 2: Use the ideal gas law (PV=nRT) to find the number of moles of H2 gas.

n = PV/RT n = (0.80789 atm * 0.76244 L) / (0.0821 Latm/molK * 313.01 K) = 0.0237 mol H2

Step 3: Use the stoichiometry of the reaction to find the moles of Mg. According to the reaction, 1 mol of Mg produces 1 mol of H2.

So, moles of Mg = moles of H2 = 0.0237 mol Mg

Step 4: Convert moles of Mg to grams using the molar mass of Mg (24.305 g/mol).

Mass of Mg = moles of Mg * molar mass of Mg Mass of Mg = 0.0237 mol Mg * 24.305 g/mol = 0.576 g Mg

So, you would need approximately 0.576 grams of Mg to produce 762.44 mL of H2 gas under the given conditions.