0.042g of Magnesium was reacted with excess HCl at S.T.P.
A. Write the equation for the reaction
B. How many moles of Mg were used?
C. What volume of Hydrogen is produced in this reaction?_
Mg + 2HCl = MgCl2 + H2
.042g Mg = .00173 mole Mg
so, you get .00173 mol H2
it will occupy .00173*22.4 = .0387L, or 38.7mL
A. The equation for the reaction between magnesium and hydrochloric acid is as follows:
Mg + 2HCl -> MgCl2 + H2
B. To calculate the number of moles of magnesium used, we can use the given mass of magnesium and its molar mass. The molar mass of magnesium is 24.31 g/mol.
First, we convert the given mass of magnesium to moles:
moles of Mg = mass of Mg / molar mass of Mg
moles of Mg = 0.042 g / 24.31 g/mol
moles of Mg ≈ 0.00173 mol
Therefore, 0.00173 moles of magnesium were used in the reaction.
C. To find the volume of hydrogen produced, we need to use the concept of stoichiometry. The balanced equation shows that 1 mole of magnesium produces 1 mole of hydrogen. At STP (standard temperature and pressure), 1 mole of a gas occupies 22.4 liters.
Hence, the volume of hydrogen gas produced can be calculated as follows:
volume of H2 = number of moles of H2 * molar volume of gas at STP
volume of H2 = 0.00173 mol * 22.4 L/mol
volume of H2 ≈ 0.0387 L or 38.7 mL
Therefore, approximately 0.039 liters (or 39 mL) of hydrogen gas is produced in this reaction.