Question

Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 9.9 g of Pb(NO3)2are heated to give 5.5 g of PbO??

Answer 1

HEY MATE HERE IS YOUR ANSWER:-

We need (i) a stoichiometric equation........

We need (i) a stoichiometric equation........Pb(NO3)2+Δ→PbO(s)+2NO2(g)+12O2(g)

We need (i) a stoichiometric equation........Pb(NO3)2+Δ→PbO(s)+2NO2(g)+12O2(g)This simply must be known; nitrates undergo thermal decomposition to give oxide and nitric oxide and dioxygen gas.

We need (i) a stoichiometric equation........Pb(NO3)2+Δ→PbO(s)+2NO2(g)+12O2(g)This simply must be known; nitrates undergo thermal decomposition to give oxide and nitric oxide and dioxygen gas.And (ii). given the 1:1 stoichiometry between nitrate and oxide, we needs equivalent quantities of lead oxide and lead nitrate:

We need (i) a stoichiometric equation........Pb(NO3)2+Δ→PbO(s)+2NO2(g)+12O2(g)This simply must be known; nitrates undergo thermal decomposition to give oxide and nitric oxide and dioxygen gas.And (ii). given the 1:1 stoichiometry between nitrate and oxide, we needs equivalent quantities of lead oxide and lead nitrate:Percentage yield=Moles of lead oxide÷

We need (i) a stoichiometric equation........Pb(NO3)2+Δ→PbO(s)+2NO2(g)+12O2(g)This simply must be known; nitrates undergo thermal decomposition to give oxide and nitric oxide and dioxygen gas.And (ii). given the 1:1 stoichiometry between nitrate and oxide, we needs equivalent quantities of lead oxide and lead nitrate:Percentage yield=Moles of lead oxide÷Moles of lead nitrate×100%

We need (i) a stoichiometric equation........Pb(NO3)2+Δ→PbO(s)+2NO2(g)+12O2(g)This simply must be known; nitrates undergo thermal decomposition to give oxide and nitric oxide and dioxygen gas.And (ii). given the 1:1 stoichiometry between nitrate and oxide, we needs equivalent quantities of lead oxide and lead nitrate:Percentage yield=Moles of lead oxide÷Moles of lead nitrate×100%=5.5g ÷ 223.2g mol.−1 ÷ 9.9 g ÷ 331.2gmol−1×100%

HOPE IT HELP YOU..

Answer 2

The percent yield of the reaction is 82.46 %

Explanation:

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

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

Given mass of lead (II) nitrate = 9.9 g

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

Putting values in equation 1, we get:

$\text{Moles of lead (II) nitrate}=\frac{9.9g}{331.2g/mol}=0.0299mol$

The chemical equation for the decomposition of lead (II) nitrate follows:

$2Pb(NO_3)_2\rightarrow 2PbO+4NO_2+O_2$

By Stoichiometry of the reaction:

2 moles of lead (II) nitrate produces 2 moles of lead (II) oxide

So, 0.0299 moles of lead (II) nitrate will produce = $\frac{2}{2}\times 0.0299=0.0299mol$ of lead (II) oxide

Now, calculating the mass of lead (II) oxide from equation 1, we get:

Molar mass of lead (II) oxide = 223.2 g/mol

Moles of lead (II) oxide = 0.0299 moles

Putting values in equation 1, we get:

$0.0299mol=\frac{\text{Mass of lead (II) oxide}}{223.2g/mol}\\\\\text{Mass of lead (II) oxide}=(0.0299mol\times 223.2g/mol)=6.67g$

To calculate the percentage yield of lead (II) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of lead (II) oxide = 5.5 g

Theoretical yield of lead (II) oxide = 6.67 g

Putting values in above equation, we get:

$\%\text{ yield of lead (II) oxide}=\frac{5.5g}{6.67g}\times 100\\\\\% \text{yield of lead (II) oxide}=82.46\%$

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