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lozanna [386]
3 years ago
14

I need help with the practice question at the bottom. Thank you.

Chemistry
1 answer:
algol133 years ago
6 0

Explanation:

For this question, we apply the equation: Q = mCp AT Where m is the mass of the substance, Cp

is its specific heat capacity and AT is the

temperature change. Q = 896 x 0.45 x (5-94)

Q = -35884.8 Joules

So about -36 kilojoules of heat is released.

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04.05 mol
victus00 [196]

Answer:

34.8 g

Explanation:

Answer:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ:         123.90    70.91      208.24

                 P₄   +   20Cl₂ ⟶ 4PCl₅

Mass/g:   46.0       32.0

2. Calculate the moles of each reactant  

\text{moles of P}_{4} = \text{46.0 g P}_{4} \times \dfrac{\text{1 mol P}_{4}}{\text{123.90 g P}_{4}} = \text{0.3713 mol P}_{4}\\\\\text{moles of Cl}_{2} = \text{32.0 g Cl }_{2} \times \dfrac{\text{1 mol Cl }_{2}}{\text{70.91 g Cl }_{2}} = \text{0.4513 mol Cl }_{2}

3. Calculate the moles of PCl₅ we can obtain from each reactant

From P₄:

The molar ratio is 4 mol PCl₅:4 mol P₄

\text{Moles of PCl}_{5} = \text{0.3713 mol P}_{4} \times \dfrac{\text{4 mol PCl}_{5}}{\text{4 mol P}_{4}} = \text{0.3713 mol PCl}_{5}

From Cl₂:

The molar ratio is 4 mol PCl₅:20 mol Cl₂

\text{Moles of PCl}_{5} =  \text{0.4513 mol Cl}_{2}\times \dfrac{\text{4 mol PCl}_{5}}{\text{20 mol Cl}_{2}} = \text{0.090 26 mol PCl}_{5}

4. Identify the limiting and excess reactants

The limiting reactant is chlorine, because it gives the smaller amount of PCl₅.

The excess reactant is phosphorus.

5. Mass of excess reactant

(a) Moles of P₄ reacted

The molar ratio is 1 mol P₄:20 mol Cl₂

\text{Moles reacted} = \text{0.4513 mol Cl}_{2} \times \dfrac{\text{4 mol P}_{4}}{\text{20 mol Cl}_{2}} = \text{0.090 26 mol P}_{4}

(b) Mass of P₄ reacted

\text{Mass reacted} = \text{0.090 26 mol P}_{4} \times \dfrac{\text{123.90 g P}_{4}}{\text{1 mol P}_{4}} = \text{11.18 g P}_{4}

(c) Mass of P₄ remaining

Mass remaining = original mass – mass reacted = (46.0 - 11.18) g = 34.8 g P₄

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