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Considering this balanced chemical equation, how many grams of HgO will be produced when 44 g of Hg react with excess O2?2Hg (l) + O2 (g)→ 2HgO (s) Group of answer choices48 g96 g44 g28 g

Question

Considering this balanced chemical equation, how many grams of HgO will be produced when 44 g of Hg react with excess O2?2Hg (l) + O2 (g)→ 2HgO (s) Group of answer choices48 g96 g44 g28 g

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Solution

To solve this problem, we need to use stoichiometry, which is a method in chemistry that uses the relationships between reactants and products in a chemical reaction to calculate desired quantitative data.

Here are the steps:

  1. First, we need to know the molar mass of Hg (Mercury), which is approximately 200.59 g/mol.

  2. Then, we calculate the number of moles of Hg in 44 g. We use the formula: Number of moles = mass / molar mass So, the number of moles of Hg = 44 g / 200.59 g/mol = 0.219 moles

  3. According to the balanced chemical equation, the ratio of Hg to HgO is 1:1. This means that 1 mole of Hg will produce 1 mole of HgO.

  4. Therefore, 0.219 moles of Hg will produce 0.219 moles of HgO.

  5. Finally, we convert the moles of HgO back to grams. The molar mass of HgO (Mercury(II) oxide) is approximately 216.59 g/mol. So, the mass of HgO = number of moles * molar mass Mass of HgO = 0.219 moles * 216.59 g/mol = 47.4 g

So, approximately 48 g of HgO will be produced when 44 g of Hg react with excess O2.

This problem has been solved

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