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navik [9.2K]
3 years ago
10

What is the limiting reactant if 8 g of Ba reacts with 2.8 g of Al2(SO4)3?

Chemistry
1 answer:
Nata [24]3 years ago
5 0

Answer:

Al2(SO4)3 is the limiting reactant

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3Ba + Al2(SO4)3 → 2Al + 3BaSO4

Next, we shall determine the mass of Ba and the mass of Al2(SO4)3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Al2(SO4)3 = 2x27 + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96]

= 54 + 288 = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Therefore, 8g of Ba will react with

= (8 x 342/411 = 6.66g of Al2(SO4)3.

From the calculations made above, we can see that it will take a higher mass of Al2(SO4)3 i.e 6.66g than what was given i.e 2.8g to react completely with 8g of Ba.

Therefore, Al2(SO4)3 is the limiting reactant and Ba is the excess reactant.

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Refer to the Chemical Compounds table above. How many total elements are in<br> Ethanol?
Korolek [52]

Answer:

3 element i.e carbon (C), hydrogen (H) and oxygen.

Explanation:

The following data were obtained from the question:

Subtance >>>>>>>> Chemical Formula

Glucose >>>>>>>>> C₆H₁₂O₆

Methane >>>>>>>> CH₄

Ethanol >>>>>>>>> C₂H₅OH

Hydrogen peroxide >> H₂O₂

From the above table, we can see that ethanol (C₂H₅OH) contains carbon (C), hydrogen (H) and oxygen

Therefore, the total number of elements present in ethanol, C₂H₅OH is 3.

8 0
3 years ago
What would the products be for the reaction between Na3PO4 + MgSO4?
ivann1987 [24]

MgSO4 + Na3PO4 = Na2SO4 + Mg3(PO4)2

Answer: The products of Na3PO4 + MgSO4 are Na2SO4 + Mg3(PO4)2

Explanation:

7 0
3 years ago
38.25 grams of silicon is combined with 14.33 grams of nitrogen gas. How many grams of silicon nitride can be formed if nitrogen
zhuklara [117]
3Si + 2N2 --> Si3N4 (as given) 

n(Si) = m/MM = 38.25/28.085 = 1.3619 mol
n(N2) = 14.33/2*14.007 = 0.5115 mol

Therefore, N2 is limiting and Si is in excess 
The molar ratio of 2N2:Si3N4 is 2:1 
So, 0.0575 mol of silicon nitride is formed (dividing 0.5115 by 2) 

m of silicon nitride= n*mm = 0.0575*140.283 = 8.06627... g 
= 8.066g (4 significant figures) 

(hopefully it is right, but double check in case i did something wrong) :) 
6 0
3 years ago
Which metal will react spontaneously with Cu2+ (aq) at 25°C?
Gwar [14]

Answer:

Mg  

Explanation:

The standard reduction potentials are

                                              <u>E°/V </u>

Au³⁺(aq ) + 3e⁻  ⟶  Au(s);     1.42

Hg²⁺(aq)  + 2e⁻  ⟶  Hg(l);     0.85

Ag⁺(aq)    +   e⁻  ⟶  Ag(s);    0.80

Cu²⁺(aq)   + 2e⁻ ⟶  Cu(s);   0.34

Mg2+(aq) + 2e- ⟶  Mg(s);   -2.38

The more negative the standard reduction potential, the stronger the metal is as a reducing agent.

Mg is the only metal with a standard reduction potential lower than that of Cu, so

Only Mg will react spontaneously with Cu²⁺.

 

5 0
3 years ago
Read 2 more answers
Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---&gt; C2H2(g) ΔH° = ??? kJ Given the following thermochemical equation
nordsb [41]

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

C_2H_2(g) + \frac{5}{2}O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_{1} = -1,123 kJ...[1]

C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_{2} = -340 kJ..[2]

H_2(g) + \frac{1}{2}O_2(g)\rightarrow H_2O(l) ,\Delta H^o_{3} = -211 kJ..[3]

2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_{4} =?..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

\Delta H^o_{4}=2\times \Delta H^o_{2}+\Delta H^o_{3} - \Delta H^o_{1}

\Delta H^o_{4}=2\times (-340 kJ) + (-211 kJ) - (-1,123 kJ)

\Delta H^o_{4}=232 kJ

The enthalpy for given reaction is 232 kilo Joules.

5 0
3 years ago
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