Write a balanced chemical equation for the reaction of magnesium with hydrochlor

Question

Write a balanced chemical equation for the reaction of magnesium with hydrochloric acid. A class ran this lab a described above. a. One student cut a piece of magnesium ribbon and found that it weighed 0.0347 g. The magnesium reacted with the 4.0 mL of concentrated hydrochloric acid. The concentration of concentrated hydrochloric acid is 12 mol/liter. Which reactant is the limiting reactant, the magnesium or the hydrochloric acid? b. How many moles of hydrogen gas will this piece of magnesium produce when reacted with 4 mL of 12 M hydrochloric acid? c. How many mL of hydrogen gas at STP would you expect to be produced by the reaction in b? d. What gases were in the eudiometer tube above the water after the reaction was complete? e. The barometric pressure was measured as 752 mm of Hg and the column of water in the eudiometer tube was 34 mm tall. What was the pressure of the gas mixture in mm of Hg? The temperature of the solution in the 800 ml beaker was 24 degree C. What was the pressure of the hydrogen gas in the eudiometer tube in mm of Hg?

Explanation / Answer

Given

Molarity of HCl = 12 M (mol/L)

Volume of HCl = 4 ml = 0.004 L

No. of moles of HCl = 0.004 L * 12 mol/L = 0.048 moles of HCl

from the previous question

mass of magnesium = 0.037 g

Molar mass of magnesium = 24.3 g/mol

No. of moles of magnesium = mass/ molar mass = 0.037 g / 24.3 g/mol = 0.00152 moles

According to given reaction stoichometry

1 mole of Mg requires 2 mole of HCl

so 0.00152 moles of Mg requires 0.00304 moles of HCl so Mg is the limiting reactant

Accordign to reaction stoichometry

1 mol of Mg will give 1 mol of H2

No. of moles of H2 produced = 0.00152 moles Answer (b)

Given at STP

at STP T = 0 C = 273 K P = 1 atm R = 0.08206 L.atm/mol.K

P*V = n*R*T

1 atm * V = 0.00152 moles * 0.08206 L.atm/mol.K * 273 K

V = 0.034 L Answer (c)

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