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2 years ago

Need help please?!!!!!

Chemistry

1 answer:

puteri [66]2 years ago

6 0

(a) 33.6 L of oxygen would be produced.

(b) 106 grams of $Na_2CO_3$ would be needed

<h3>Stoichiometric calculations</h3>

1 mole of gas = 22.4 L

(a) From the equation, 2 moles of $KClO_3$ produce 3 moles of $O_2$ . 1 mole of $KClO_3$ will, therefore, produce 1.5 moles of $O_2$ .

1.5 moles of oxygen = 22.4 x 1.5 = 33.6 L

(b) 22.4 L of $CO_2$ is produced at STP. This means that 1 mole of the gas is produced.

From the equation, 1 mole of $CO_2$ requires 1 mole of $Na_2CO_3$ .

Molar mass of $Na_2CO_3$ = (23x2)+ (12)+(16x3) = 106 g/mol

Mass of 1 mole $Na_2CO_3$ = 1 x 106 = 106 grams

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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I NEED HELP PLEASE, THANKS! :)

krok68 [10]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

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

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

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

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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Need help please?!!!!!​

Need help please?!!!!!