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The equation for the reaction of magnesium with hydrochloric acid is Mg(s) + 2hcl(aq) MgCl2(aq) + H2(g) calculate the relative formula masses (m_{r}) of: Calculate the volume of hydrogen (at 1 atmosphere and 25 deg * c ) which could be obtained from 1 g of magnesium reacting with excess hydrochloric acid.

Accepted Answer
1. Calculate the relative formula masses (M_r):

Mg: 24.31 g/mol
HCl: 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46 g/mol
MgCl2: 24.31 g/mol (Mg) + 2
35.45 g/mol (Cl) = 95.21 g/mol
H2: 2
1.01 g/mol (H) = 2.02 g/mol
2. Calculate the moles of magnesium:

Moles of Mg = mass / M_r = 1 g / 24.31 g/mol = 0.0412 mol
3. Calculate the moles of hydrogen produced:

From the balanced equation, 1 mole of Mg produces 1 mole of H2. Therefore, 0.0412 mol of Mg will produce 0.0412 mol of H2.
4. Calculate the volume of hydrogen using the ideal gas law:

PV = nRT
Where:

P = pressure = 1 atm

V = volume (what we want to find)

n = moles of H2 = 0.0412 mol

R = ideal gas constant = 0.0821 L atm/mol K

T = temperature = 25°C = 298 K

V = nRT / P = (0.0412 mol
0.0821 L atm/mol K
298 K) / 1 atm = 1.01 L
Therefore, 1 g of magnesium reacting with excess hydrochloric acid will produce approximately 1.01 liters of hydrogen gas at 1 atmosphere and 25°C.

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