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photoshop1234 [79]
3 years ago
11

Sodium metal reacts with chlorine gas to produce sodium chloride. What mass (grams) of sodium metal would be needed to fully rea

ct with 2.3 moles of chlorine gas?
Chemistry
1 answer:
gulaghasi [49]3 years ago
3 0

Answer:

105.8 g of Na would be required

Explanation:

Let's think the reaction:

2Na(s)  + Cl₂(g)  →  2NaCl (s)

1 mol of chlorine reacts with 2 moles of sodium

Then, 2.3 moles of Cl₂ would react with (2.3 .2) / 1 = 4.6 moles

Let's determine the mass of them.

4.6 mol . 23 g/mol = 105.8 g

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Which of the following best explains the polarity of water (H2O)?
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Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 4.90 g of su
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Answer:

Mass H2SO4 = 3.42 grams

Mass of lead acetate = 0 grams

Mass PbSO4 = 4.58 grams

Mass of CH3COOH = 1.81 grams

Explanation:

Step 1: Data given

Mass of sulfuric acid = 4.90 grams

Molar mass of sulfuric acid = 98.08 g/mol

Mass of lead acetate = 4.90 grams

Molar mass of lead acetate = 325.29 g/mol

Step 2: The balanced equation

H2SO4 + Pb(C2H3O2)2 → PbSO4 + 2CH3COOH

Step 3: Calculate moles

Moles = mass / molar mass

Moles H2SO4 = 4.90 grams / 98.08 g/mol

Moles H2SO4 = 0.0500 moles

Moles lead acetate = 4.9 grams / 325.29 g/mol

Moles lead acetate = 0.0151 moles

Step 4: Calculate the limiting reactant

For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH

The limiting reactant is lead acetate. It will completzly be consumed (0.0151 moles). H2SO4 is in excess. There will react 0.0151 moles. There will remain 0.0500 - 0.0151 = 0.0349 moles

Step 5: Calculate moles of products

For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH

For 0.0151 moles lead acetate we'll have 0.0151 moles PbSO4 and 2*0.0151 = 0.0302 moles CH3COOH

Step 6: Calculate mass

Mass = moles * molar mass

Mass H2SO4 = 0.0349 moles * 98.08 g/mol

Mass H2SO4 = 3.42 grams

Mass PbSO4 = 0.0151 moles * 303.26 g/mol

Mass PbSO4 = 4.58 grams

Mass of CH3COOH = 0.0302 moles * 60.05 g/mol

Mass of CH3COOH = 1.81 grams

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