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ivanzaharov [21]
3 years ago
8

2NaClO3 ——> 2NaCl + 3O2

Chemistry
1 answer:
Alborosie3 years ago
3 0

Answer:  

<h3>1) 208.09 g</h3><h3>2) 7.11 × 10²² particles</h3>

<h2>Explanation:</h2><h3>NUMBER ONE</h3>

<u>Find moles of Known Substance (NaCl)</u>

Moles of NaCl = mass ÷ molar mass

                        =  114.25 g ÷ 58.44 g/mol

                        = 1.955 mol

<u>Use Mole Ratio of Known to Unknown (NaCl : NaClO₃) to Determine Moles of Unknown</u>

The mole ratio of NaCl to NaClO₃ is 2 to 2 or simply 1 to 1.

This means that for each mole of NaCl produced the same number of moles of NaClO₃ was decomposed.

∴ since moles of NaCl = 1.955 mol,

then moles of NaClO₃ = 1.955 mol

<u>Find Mass of Unknown </u>

Mass of NaClO₃ = moles × molar mass

                           = 1.955 mol × 106.44 g/mol

                           =  208.09 g

<h3>NUMBER TWO</h3>

<u>Determine the Moles of Known (O₂)</u>

moles of O₂ = (volume × density) ÷ molar mass

                    = (68.30 L × 0.0828 g/L) ÷ 32 g/mol

                    = 0.177 mol

<u>Use Mole Ratio of Known to Unknown (O₂ : NaClO₃) to Determine Moles of Unknown</u>

Mole ratio of O₂ : NaClO₃ is 3 : 2. Therefore for each mole of oxygen produced, ²/₃ the moles of NaClO₃ is produced.

∴ if moles of O₂  =  0.177 mol

then moles of NaClO₃ =  0.177 mol × ²/₃

                                     = 0.118 mol

<u>Find the Number of Particles of Unknown</u>

Number of particles  = moles × Avogadro's Number

                                  = 0.118 mol × (6.022 × 10²³ particles/mol)

                                  =  7.11 × 10²² particles

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3 years ago
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The mass of sodium chloride at the two parts are mathematically given as

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<h3>What is the mass of sodium chloride that can react with the same volume of fluorine gas at STP?</h3>

Generally, the equation for ideal gas is mathematically given as

PV=nRT

Where the chemical equation is

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