From the mass of Mg provided, calculate the volume of hydrogen gas that should have been produced. Use 100.0 kPA as the barometric pressure. (You will need to use DA, gas stoich, Dalton’s Law and PV = nRT ). * Use average values for calculations ### Equations**Molecular Equation**: \[ \text{2HCl (aq) + Mg (s) → H}_2 \text{ (g) + MgCl}_2 \text{ (aq)} \]**Net Ionic Equation**: \[ \text{2H}^+ \text{ (aq) + Mg (s) → H}_2 \text{ (g) + Mg}^{2+} \text{ (aq)} \]### Data| | Trial One | Trial Two | Average ||----------------------------|-----------|-----------|---------|| **Mass of Mg Strip (g)** | 0.0342 g | 0.0350 g | 0.0346 g || **Volume of H\(_2\) at room temperature (mL)** | 36.5 mL | 37.4 mL | 36.9 mL |**Description:**This data table presents the results of two trials involving a chemical reaction between hydrochloric acid (HCl) and magnesium (Mg). The reaction produces hydrogen gas (H\(_2\)) and magnesium chloride (MgCl\(_2\)).- **Mass of Mg Strip:** The mass of the magnesium strip used in each trial is recorded, with values of 0.0342 g and 0.0350 g for Trial One and Trial Two, respectively. The average mass is calculated to be 0.0346 g. - **Volume of H\(_2\):** The volume of hydrogen gas collected at room temperature is measured in milliliters (mL), yielding results of 36.5 mL for Trial One and 37.4 mL for Trial Two. The average volume is 36.9 mL.These results give insights into the consistency and accuracy of the measurements during the experiment.

The reaction of metal with concentrated acids liberates hydrogen gas. The quantity of the hydrogen gas produced depends on the mass of the metal reacted if the acid is taken in excess. Once the moles of the limiting agent (that consumed completely in a chemical reaction) are determined, the stoichiometric equation is related to evaluate the moles of hydrogen gas produced. The gas equation that relates pressure (P), volume (V), temperature (T), and the number of moles (n) are expressed as V=nRTP ...(1). The average mass (mA) of the Mg is determined as follows: mA=(0.0342+0.0350+0.0346) g3=0.0346 g The average volume (VH) of hydrogen gas produced is determined as follows: VH=(36.5+37.4+36.9) mL3=36.9 mL Calculate the number of moles of Mg (nMg) reacted as follows: nMg=0.0346 g24.305 g/mol=0.00142 mol From the balanced equation, it is observed that one mole of Mg produces one mole of hydrogen gas. So, 0.00142 mol of Mg produces 0.00142 mol of hydrogen gas. At temperature (T) of 25°C, pressure (P) of 100.0 kPA, the volume of hydrogen gas produced (V) is determined as follows: P=100.0 kPa=0.00986 atm1 kPa×100.0 kPa=0.986 atmT=25°C=(25+273) K=298 KV=(0.00142 mol)(0.0820 L atm/K mol)(298 K)0.986 atm=0.0352 L=35.19 mLHence the volume of the hydrogen gas produced is computed as 35.19 mL.