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Furkat [3]
3 years ago
12

3.0 g aluminum and 6.0 g of bromine react to form AlBr3 2Al+3Br2=2AlBr3 How much product would be produced? How much reagent wou

ld remain at the end of this reaction?
Chemistry
1 answer:
juin [17]3 years ago
8 0
<span>2Al + 3Br2 -------------> 2AlBr3

</span>3 g Al = 0.11 mol Al. 

<span>6 g Br2 = 0.0375 mole bromine (it is diatomic). </span>

<span>moles of aluminium will take part in reaction = 0.0375 X (2/3) = 0.025. </span>

<span>Gram-mole of AlBr3 will be produced = 0.025 mole = 6.6682 g. 
</span>moles of Al left = 0.11 - 0.025 = 0.086<span>


</span>
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padilas [110]

Answer:

0.143 g of KCl.

Explanation:

Equation of the reaction:

AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)

Molar concentration = mass/volume

= 0.16 * 0.012

= 0.00192 mol AgNO3.

By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.

Number of moles of KCl = 0.00192 mol.

Molar mass of KCl = 39 + 35.5

= 74.5 g/mol

Mass = molar mass * number of moles

= 74.5 * 0.00192

= 0.143 g of KCl.

4 0
3 years ago
Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th
WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]

We are given:

\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

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3 years ago
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How many total moles of ions are released when the following sample dissolves completely in water? 6.188 x 10 21 formula units o
elixir [45]

Answer:

e. 3.08 x 10⁻² mol of ions.

Explanation:

  • Every 1.0 mole of any compound contains Avogadro's number of molecules (6.022 x 10²³).

  • We can get the no. of moles of NiCl₂ using cross multiplication:

1.0 mol NiCl₂ contains → 6.022 x 10²³ molecules.

??? mol NiCl₂ contains → 6.188 x 10²¹ molecules.

∴ The no. of moles of NiCl₂ = (1.0 mol)(6.188 x 10²¹ molecules)/(6.022 x 10²³ molecules) = 1.028 x 10⁻² mol.

  • NiCl₂ is ionized according to the equation:

NiCl₂ → Ni²⁺ + 2Cl⁻.

Which means that every 1.0 mol of NiCl₂ is ionized to produce 3.0 moles (1.0 mol of Ni²⁺ and 2 moles of Cl⁻).

<em>∴ The total moles of ions are released</em> = 3 x 1.028 x 10⁻² mol = <em>3.083 x 10⁻² mol of ions.</em>

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