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jok3333 [9.3K]
2 years ago
13

Iron has a density of 7.9 g/cm3. The volume of a regular cylinder is V =πr2h. An iron cylinder has a height of 3.00 m and a mass

of 2.00 kg. Using the value 3.1416 for the constant π, the radius in cm of the iron cylinder is?
Chemistry
1 answer:
Sedaia [141]2 years ago
6 0
Hfhsnebrjicjcbdnd hope this helps
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\large \boxed{\text{2.20 g Pb}}

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They gave us the masses of two reactants and asked us to determine the mass of the product.

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Mᵣ:       239.27   32.00        207.2

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\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

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