QuestionMay 19, 2026

b. How many particles of NaClO_(3) are needed to produce 68.30 g oxygen? underline ( )Xunderline ( )Xunderline ( )Xunderline ( )

b. How many particles of NaClO_(3) are needed to produce 68.30 g oxygen? underline ( )Xunderline ( )Xunderline ( )Xunderline ( )
b. How many particles of NaClO_(3) are needed to produce 68.30 g oxygen? underline ( )Xunderline ( )Xunderline ( )Xunderline ( )

Solution
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Answer

8.56 \times 10^{23} particles Explanation 1. Write and balance the chemical equation 2NaClO_{3} \rightarrow 2NaCl + 3O_{2} 2. Find moles of O_{2} Molar mass of O_{2} = 32.00 \, g/mol n(O_{2}) = \frac{68.30}{32.00} = 2.134375 \, mol 3. Use mole ratio (NaClO_{3} to O_{2}) From the equation: 2 \, mol \, NaClO_{3} \rightarrow 3 \, mol \, O_{2} n(NaClO_{3}) = 2.134375 \times \frac{2}{3} = 1.4229167 \, mol 4. Convert moles to number of particles **N = n \times N_A**, where N_A = 6.022 \times 10^{23} N = 1.4229167 \times 6.022 \times 10^{23} = 8.564 \times 10^{23} particles

Explanation

1. Write and balance the chemical equation<br /> $2NaClO_{3} \rightarrow 2NaCl + 3O_{2}$<br /><br />2. Find moles of $O_{2}$<br /> Molar mass of $O_{2} = 32.00 \, g/mol$ <br /> $n(O_{2}) = \frac{68.30}{32.00} = 2.134375 \, mol$<br /><br />3. Use mole ratio ($NaClO_{3}$ to $O_{2}$)<br /> From the equation: $2 \, mol \, NaClO_{3} \rightarrow 3 \, mol \, O_{2}$ <br /> $n(NaClO_{3}) = 2.134375 \times \frac{2}{3} = 1.4229167 \, mol$<br /><br />4. Convert moles to number of particles<br /> **$N = n \times N_A$**, where $N_A = 6.022 \times 10^{23}$ <br /> $N = 1.4229167 \times 6.022 \times 10^{23} = 8.564 \times 10^{23}$ particles
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