QuestionDecember 17, 2025

What mass of carbon (in g) is present in 8.02times 10^23 formula units of Al_(2)(CO_(3))_(3) square

What mass of carbon (in g) is present in 8.02times 10^23 formula units of Al_(2)(CO_(3))_(3) square
What mass of carbon (in g) is present in 8.02times 10^23 formula units of Al_(2)(CO_(3))_(3) square

Solution
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Answer

48.04\ \text{g} Explanation 1. Determine moles of formula units Use **n = \frac{N}{N_A}**, with N_A = 6.022\times 10^{23} mol⁻¹. n = \frac{8.02\times 10^{23}}{6.022\times 10^{23}} \approx 1.33 mol of Al_{2}(CO_{3})_{3}. 2. Find moles of carbon Each formula unit has 3 CO₃ groups → 3 C atoms/group → total 3 × 1 = 3 C atoms per formula unit. Moles of C = 1.33 \times 3 \approx 4.00 mol C. 3. Convert moles of carbon to mass **m = n \times M**, with M_C = 12.01 g/mol. m = 4.00 \times 12.01 \approx 48.04 g.

Explanation

1. Determine moles of formula units <br /> Use **$n = \frac{N}{N_A}$**, with $N_A = 6.022\times 10^{23}$ mol⁻¹. <br />$n = \frac{8.02\times 10^{23}}{6.022\times 10^{23}} \approx 1.33$ mol of $Al_{2}(CO_{3})_{3}$. <br /><br />2. Find moles of carbon <br /> Each formula unit has 3 CO₃ groups → 3 C atoms/group → total 3 × 1 = 3 C atoms per formula unit. <br />Moles of C = $1.33 \times 3 \approx 4.00$ mol C. <br /><br />3. Convert moles of carbon to mass <br /> **$m = n \times M$**, with $M_C = 12.01$ g/mol. <br />$m = 4.00 \times 12.01 \approx 48.04$ g.
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