30.03.2023 - 18:11

Magnesium metal is added to 6M hydrochloric acid and reacts according to the following balanced equation: Mg(s) + 2 HCI(aq) rightarrow MgCl_2 (aq) + H_2(g) a. What volume of 1.8 M hydrochloric acid will react with 22.1 grams of Mg? b. What mass of magnes

Question:

Magnesium metal is added to 6M hydrochloric acid and reacts according to the following balanced equation:

{eq}Mg(s) + 2 HCI(aq) rightarrow MgCl_2 (aq) + H_2(g) {/eq}

a. What volume of 1.8 M hydrochloric acid will react with 22.1 grams of Mg?

b. What mass of magnesium will react with 1.75 liters of 0.4 M hydrochloric acid?

c. How many grams of magnesium chloride will you make (at STP) if you react 1.5 L of 3.0 M hydrochloric acid with excess magnesium?

Answers (1)
  • Freda
    April 5, 2023 в 10:38
    a. To calculate the volume of 1.8 M hydrochloric acid that will react with 22.1 grams of Mg, we first need to use stoichiometry to determine how much hydrochloric acid is required to react with 1 mole of magnesium. From the balanced chemical equation, we see that 1 mole of Mg reacts with 2 moles of HCl. The molar mass of Mg is 24.31 g/mol, so 22.1 g of Mg is equivalent to 0.91 moles of Mg. Therefore, we need 1.82 moles of HCl to react with this amount of Mg (since the mole ratio of Mg to HCl is 1:2). Next, we can use the formula for molarity to calculate the volume of 1.8 M hydrochloric acid that contains 1.82 moles of HCl: Molarity = moles / volume Solving for volume, we get: volume = moles / molarity = 1.82 / 1.8 = 1.01 L Therefore, we need 1.01 liters of 1.8 M hydrochloric acid to react with 22.1 grams of Mg. b. To calculate the mass of magnesium that will react with 1.75 liters of 0.4 M hydrochloric acid, we again need to use stoichiometry to determine the amount of Mg required to react with this amount of HCl. From the balanced chemical equation, we see that 1 mole of Mg reacts with 2 moles of HCl. The molarity of the HCl solution is 0.4 M, which means it contains 0.4 moles of HCl per liter. Therefore, the 1.75 liters of 0.4 M HCl contains: 0.4 moles / L x 1.75 L = 0.7 moles of HCl Since the mole ratio of Mg to HCl is 1:2, we need half as many moles of Mg as HCl to ensure complete reaction. Therefore, we need 0.35 moles of Mg to react with 0.7 moles of HCl. The molar mass of Mg is 24.31 g/mol, so 0.35 moles of Mg is equivalent to: 0.35 moles x 24.31 g/mol = 8.51 g of Mg Therefore, 8.51 grams of Mg will react with 1.75 liters of 0.4 M hydrochloric acid. c. To calculate the mass of magnesium chloride produced when 1.5 L of 3.0 M hydrochloric acid is reacted with excess magnesium, we first need to determine the number of moles of HCl in 1.5 L of 3.0 M HCl: Molarity = moles / volume Solving for moles, we get: moles = molarity x volume = 3.0 x 1.5 = 4.5 moles of HCl Since the mole ratio of Mg to HCl is 1:2, we need twice as many moles of HCl as Mg to ensure complete reaction. Therefore, we need 2.25 moles of Mg to react with 4.5 moles of HCl. The molar mass of Mg is 24.31 g/mol, so 2.25 moles of Mg is equivalent to: 2.25 moles x 24.31 g/mol = 54.6 g of Mg From the balanced chemical equation, we see that 1 mole of MgCl2 is produced for every mole of Mg that reacts: Mg(s) + 2 HCl(aq) ? MgCl2(aq) + H2(g) Therefore, 2.25 moles of MgCl2 will be produced when 2.25 moles of Mg reacts with excess HCl. The molar mass of MgCl2 is 95.21 g/mol, so 2.25 moles of MgCl2 is equivalent to: 2.25 moles x 95.21 g/mol = 214 g of MgCl2 Therefore, you will produce 214 grams of magnesium chloride at STP when you react 1.5 L of 3.0 M hydrochloric acid with excess magnesium.
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