What volume of 2M HCl would be required to completly react with 0.314g of Zn? what mass of hydrogen gas would be produced?
To find the volume of 2M HCl required to completely react with 0.314g of Zn, we need to use the concept of stoichiometry and balance the chemical equation between zinc (Zn) and hydrochloric acid (HCl). The balanced equation for the reaction is as follows:
Zn + 2HCl -> ZnCl2 + H2
Step 1: Calculate the number of moles (n) of Zn using its molar mass:
molar mass of Zn = 65.38 g/mol
n = mass / molar mass = 0.314g / 65.38 g/mol = 0.0048 mol
Step 2: From the balanced equation, we see that 1 mole of Zn reacts with 2 moles of HCl. Therefore, the moles of HCl required would be double the moles of Zn:
moles of HCl = 2 * 0.0048 mol = 0.0096 mol
Step 3: Calculate the volume of 2M HCl required using the equation:
volume (V) = moles (n) / concentration (C)
C = 2M (given)
V = 0.0096 mol / 2 mol/L = 0.0048 L or 4.8 mL
Therefore, the volume of 2M HCl required to completely react with 0.314g of Zn is 4.8 mL.
To find the mass of hydrogen gas produced, we need to use the stoichiometry of the balanced equation.
Step 4: The balanced equation indicates that 1 mole of Zn reacts to produce 1 mole of H2. Therefore, the moles of H2 produced would be equal to the moles of Zn reacted:
moles of H2 = 0.0048 mol
Step 5: Calculate the mass of hydrogen gas using its molar mass:
molar mass of H2 = 2 g/mol
mass of H2 = moles (n) * molar mass = 0.0048 mol * 2 g/mol = 0.0096 g or 9.6 mg
Therefore, the mass of hydrogen gas produced would be 0.0096 grams or 9.6 milligrams.