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3 years ago

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Aluminum metal reacts with aqueous nickel(II) sulfate to produce aqueous aluminum sulfate and nickel as a precipitate. In this r

eaction 108 g of aluminum were combined with 464 g of nickel(II) sulfate to produce 274 g of aluminum sulfate.

1 answer:

Marat540 [252]3 years ago

6 0

The question is incomplete, the complete question is;

In this stoichiometry problem, determine the percentage yield:

Excess aluminum metal reacts with aqueous nickel(II) sulfate to produce aqueous aluminum sulfate and nickel as a precipitate. In this reaction 108 g of aluminum were combined with 464 g of nickel(II) sulfate to produce 274 g of aluminum sulfate.

Answer:

80%

Explanation:

The reaction equation is;

2Al(s) + 3NiSO4(aq) --------> Al2(SO4)3 + 3Ni(s)

Since Al is in excess then NiSO4 is the limiting reactant.

Number of moles in 464 g of NiSO4 = mass/ molar mass

Molar mass of NiSO4 = 155 g/mol

Number of moles = 464g/155g/mol = 2.99 moles

Number of moles of Al2(SO4)3 = mass/molar mass

molar mass = 342 g/mol

Number of moles = 274g/342g/mol = 0.8 moles

From the reaction equation;

3 moles of NiSO4 yields 1 mole of Al2(SO4)3

2.99 moles of NiSO4 yields 2.99 * 1/3 = 1 mole of Al2(SO4)3

% yield = actual yield/ theoretical yield * 100/1

actual yield = 0.8 moles of Al2(SO4)3

Theoretical yield = 1 mole of Al2(SO4)3

% yield = 0.8/1 * 100 = 80%

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Vitamin C contains the elements C, H, and O. It is known to contain 40.9% C and 4.58% H by mass. The molar mass of vitamin C has

soldier1979 [14.2K]

Answer:

C₆H₈O₆

Explanation:

First off, the<u> percent of oxygen by mass</u> of vitamin C is:

100 - (40.9+4.58) = 54.52 %

<em>Assume we have one mol of vitamin C</em>. Then we would have <em>180 grams</em>, of which:

180 * 40.9/100 = 73.62 grams are of Carbon

180 * 4.58/100 = 8.224 grams are of Hydrogen

180 * 54.52/100 = 98.136 grams are of Oxygen

Now we <u>convert each of those masses to moles</u>, using the <em>elements' respective atomic mass</em>:

C ⇒ 73.62 g ÷ 12 g/mol = 6.135 mol C ≅ 6 mol C

H ⇒ 8.224 g ÷ 1 g/mol = 8.224 mol H ≅ 8 mol H

O ⇒ 98.136 g ÷ 16 g/mol = 6.134 mol O ≅ 6 mol O

So the molecular formula for vitamin C is C₆H₈O₆

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3 years ago

Limestone stalactites and stalagmites are formed in caves by the following reaction: Ca2 (aq) 2HCO−3(aq)→CaCO3(s) CO2(g) H2O(l)

Vesna [10]

Answer : The value of $\Delta E$ for this reaction is 36.18 kJ

Explanation :

First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.

As per first law of thermodynamic,

$\Delta E=q+w$

where,

$\Delta E$ = internal energy of the system

q = heat added or rejected by the system

w = work done

As we are given that:

q = 38.65 kJ

w = -2.47 kJ (system work done on surrounding)

Now put all the given values in the above expression, we get:

$\Delta E=38.65kJ+(-2.47kJ)$

$\Delta E=36.18kJ$

Therefore, the value of $\Delta E$ for this reaction is 36.18 kJ

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2 years ago

17) A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with what volume? HI

galina1969 [7]

Answer: A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

Explanation:

Given: $M_{1}$ = 0.20 M, $V_{1}$ = 15.0 mL

$M_{2}$ = 0.10 M, $V_{2}$ = ?

Formula used is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula s follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.20 M ]times 15.0 mL = 0.10 M ]times V_{2}\\V_{2} = 30 mL$

Thus, we can conclude that a 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.

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2 years ago

When 0.49 g of a molecular compound was dissolved in 20.00 g of cyclohexane, the freezing point of the solution was lowered by 3

IRISSAK [1]

Answer: The molecular mass of this compound is 131 g/mol

Explanation:

Depression in freezing point:

$\Delta T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$\Delta T_f$ = depression in freezing point = $3.9^oC$

$k_f$ = freezing point constant = $20.8^0C/m$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute = 0.49 g

$w_1$ = mass of solvent (cyclohexane) = 20.00 g

$M_2$ = molar mass of solute = ?

Now put all the given values in the above formula, we get:

$(3.9)^oC=1\times (20.8^oC/m)\times \frac{(0.49g)\times 1000}{M_2\times (20.00g)}$

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Therefore, the molar mass of solute is 131 g/mol

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2 years ago

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