The stoichiometry shows that 1 mole of Mg produces 1 mole of H2. Therefore, the number of grams of Mg required is equal to the number of moles of H2. You can multiply the moles of H2 by the molar mass of Mg to get the grams of Mg required.
To calculate the number of grams of Mg required to produce 100.00 mL of H2, we need to use the ideal gas law equation: PV = nRT.
First, we need to convert the temperature to Kelvin by adding 273.15:
T = 21.01°C + 273.15 = 294.16 K
Next, we need to convert the volume from mL to liters:
V = 100.00 mL = 0.100 L
Given that the pressure is 1.034 atm and the temperature is 294.16 K, we can rearrange the ideal gas law equation to solve for moles (n):
n = PV / RT
Substituting the values into the equation, we have:
n = (1.034 atm * 0.100 L) / (0.0821 L·atm/mol·K * 294.16 K)
Solving for n will give us the moles of H2. Since the reaction is:
Mg + 2HCl → MgCl2 + H2
The stoichiometry shows that 1 mole of Mg produces 1 mole of H2. Therefore, the number of grams of Mg required is equal to the number of moles of H2. You can multiply the moles of H2 by the molar mass of Mg to get the grams of Mg required.
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