30.03.2023 - 03:33

Mercury and bromine react with each other to produce mercury (II) bromide: Hg + Br_2 arrow HgBr_2 a) What mass of HgBr_2 can be produced from the reaction 10 g Hg and 9 g Br_2 ? b) What mass of

Question:

Mercury and bromine react with each other to produce mercury (II) bromide:

{eq}Hg + Br_2 rightarrow HgBr_2 {/eq}

a) What mass of {eq}HgBr_2 {/eq} can be produced from the reaction 10 g Hg and 9 g {eq}Br_2 {/eq}?

b) What mass of which reagent is left unreached?

c) What is the mass of {eq}HgBr_2 {/eq} can be produced form the reaction 5ml Hg (density = 13.6) and 5 ml bromine (density = 3.1)?

Answers (1)
  • Aurelia
    April 7, 2023 в 21:05
    a) To find the mass of HgBr2 produced, we must first determine which reactant limits the amount produced. This can be done using stoichiometry and comparing the moles of each reactant. From the balanced equation, we can see that 1 mole of Hg reacts with 1 mole of Br2 to produce 1 mole of HgBr2. 10 g Hg can be converted to moles of Hg by using its molar mass: 10 g Hg x (1 mol Hg/200.59 g Hg) = 0.0499 mol Hg 9 g Br2 can be converted to moles of Br2 by using its molar mass: 9 g Br2 x (1 mol Br2/159.81 g Br2) = 0.0563 mol Br2 Since the stoichiometric ratio of Hg to Br2 is 1:1, we can see that Hg is the limiting reactant, as there are fewer moles of Hg than Br2. The amount of HgBr2 produced can be calculated from the moles of Hg: 0.0499 mol Hg x (1 mol HgBr2/1 mol Hg) x (471.63 g HgBr2/1 mol HgBr2) = 23.6 g HgBr2 Therefore, 23.6 g HgBr2 can be produced from the reaction. b) To determine which reagent is left unreached, we can use the same calculations as in part a, but determine how many moles of each reactant are left over. For Hg, we know that 0.0499 mol was used up in the reaction. So the remaining moles of Hg can be found by subtracting 0.0499 from the total moles we started with: 0.0500 mol Hg - 0.0499 mol Hg = 0.0001 mol Hg For Br2, we know that 0.0563 mol was used up in the reaction. So the remaining moles of Br2 can be found by subtracting 0.0563 from the total moles we started with: 0.0563 mol Br2 - 0.0563 mol Br2 = 0 mol Br2 Therefore, all of the Br2 was used up in the reaction, and there is 0.0001 mol (or 0.0159 g) of Hg left over. c) To find the mass of HgBr2 produced, we need to first find the moles of each reactant based on their volumes and densities. For Hg: 5 mL Hg x (13.6 g/mL) = 68 g Hg 68 g Hg x (1 mol Hg/200.59 g Hg) = 0.3394 mol Hg For Br2: 5 mL Br2 x (3.1 g/mL) = 15.5 g Br2 15.5 g Br2 x (1 mol Br2/159.81 g Br2) = 0.0971 mol Br2 We can see that Hg is the limiting reactant, as there are fewer moles of Hg than Br2. So, we can use the same stoichiometric ratio as before (1:1) to calculate the moles of HgBr2 produced: 0.3394 mol Hg x (1 mol HgBr2/1 mol Hg) x (471.63 g HgBr2/1 mol HgBr2) = 160.2 g HgBr2 Therefore, 160.2 g HgBr2 can be produced from the reaction.
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