How many lead (Pb) atoms will be generated when 5.38 moles of ammonia react according to the following equation: 3PbO+2NH3→3Pb+N2+3H2O?
A) 1.62 × 10^24
B) 4.86 × 10^23
C) 2.43 × 10^24
D) 4.86 × 10^24

Answer:D. Is the correct answer

Explanation:

Following are the calculation of the lead (Pb) atoms:

Balanced reaction:  

[tex]\bold{3PbO+2NH_3\longrightarrow 3Pb+N2+3H2O}\\\\[/tex]

[tex]\bold{mol\ NH_3 \longrightarrow mol \ Pb \longrightarrow atoms \ Pb}[/tex]

Calculating the number of atoms in Pb that will be generated:

[tex]=5.38 \ mol \ NH_3 \times \frac{3\ mol\ Pb}{2 \ mol NH_3} \times \frac{6.022 \times 10^{23} \times pb}{1 \ mol \ pb}\\\\ = 4.86 \times 10^{24} \ atoms\ in \ Pb\\\\= 4.86 \times 10^{24} \ atoms \\[/tex]

Therefore, the answer is "Option D"

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