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MA_775_DIABLO [31]
4 years ago
6

According to the following balanced reaction, how many moles of HNO3 are formed from 8.44 moles of NO2 if there is plenty of wat

er present? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
Chemistry
1 answer:
kotegsom [21]4 years ago
3 0

Answer:

5.63 mol.

Explanation:

  • The balanced chemical equation between NO₂ and H₂O is:

<em>3NO₂(s) + H₂O(l) → 2HNO₃(aq) + NO(g), </em>

It is clear that 3 mol of NO₂ reacts with 1 mol of H₂O to produce 2 mol of HNO₃ and 1 mol of NO.

<em>Water is present as an excess reactant and NO₂ is limiting reactant.</em>

<em></em>

  • To find the no. of moles of HNO₃ produced:

3 mol of NO₂ produces → 2 mol of HNO₃, from stichiometry.

8.44 mol of NO₂ produces → ??? mol of HNO₃.

∴ The no. of moles of HNO₃ are formed = (8.44 mol)(2 mol)/(3 mol) = 5.63 mol.

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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
GaryK [48]

Answer:

Mass H2SO4 = 3.42 grams

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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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Explanation:

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