Question

Select the correct answer. You dissolve 8.65 grams of lead(II) nitrate in water, and then you add 2.50 grams of aluminum. This reaction occurs: 2Al(s) + 3Pb(NO3)2(aq) → 3Pb(s) + 2Al(NO3)3(aq). What’s the theoretical yield of solid lead? Use the ideal gas resource and the periodic table. A. 5.41 g B. 11.2 g C. 19.2 g D. 28.8 g

Answer 1

The amount of solid lead produced is 5.41 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For lead (II) nitrate:

Given mass of lead (II) nitrate = 8.65 g

Molar mass of lead (II) nitrate = 331.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) nitrate}=\frac{8.65g}{331.2g/mol}=0.0261mol$

• For aluminium:

Given mass of aluminium = 2.50 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{2.50g}{27g/mol}=0.0926mol$

The chemical equation for the reaction of ethylene and oxygen gas follows:

$2Al(s)+3Pb(NO_3)_2(aq.)\rightarrow 3Pb(s)+2Al(NO_3)_3(aq.)$

By Stoichiometry of the reaction:

3 moles of lead (II) nitrate reacts with 2 moles of aluminium

So, 0.0261 moles of lead (II) nitrate will react with = $\frac{2}{3}\times 0.0261=0.0174mol$ of aluminium

As, given amount of aluminium is more than the required amount. So, it is considered as an excess reagent.

Thus, lead (II) nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of lead (II) nitrate produces 3 moles of solid lead

So, 0.0261 moles of lead (II) nitrate will produce = $\frac{3}{3}\times 0.0261=0.0261moles$ of solid lead

Now, calculating the mass of lead from equation 1, we get:

Molar mass of lead = 207.2 g/mol

Moles of lead = 0.0261 moles

Putting values in equation 1, we get:

$0.0261mol=\frac{\text{Mass of lead}}{207.2g/mol}\\\\\text{Mass of lead}=(0.0261mol\times 207.2g/mol)=5.41g$

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