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3 years ago

A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant?

Chemistry

1 answer:

KengaRu [80]3 years ago

4 0

Answer:

d. Copper (II) sulfate

Explanation:

Given data:

Mass of Al = 1.25 g

Mass of CuSO₄ = 3.28 g

What is limiting reactant = ?

Solution:

Chemical equation:

2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 1.25 g/ 27 g/mol

Number of moles = 0.05 mol

Number of moles of CuSO₄:

Number of moles = mass/molar mass

Number of moles = 3.28 g/ 159.6 g/mol

Number of moles = 0.02 mol

now we will compare the moles of reactant with product.

Al : Al₂ (SO₄)₃

2 : 1

0.05 : 1/2×0.05=0.025 mol

Al : Cu

2 : 3

0.05 : 3/2×0.05 = 0.075 mol

CuSO₄ : Al₂ (SO₄)₃

3 : 1

0.02 : 1/3×0.02=0.007 mol

CuSO₄ : Cu

3 : 3

0.02 : 0.02

Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.

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Cloud [144]

Answer:

Reaction 1 is balanced but 2 is not balanced , the balance equation are :

1. $CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)$

2. $CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + H_{2}O(aq)$

Explanation:

Balanced Equations : These are the equation which follows the law of conservation of mass .

The total number of atoms present in reactant is equal to total number of atoms present in product.

1. $CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)$

This is acid - base type reaction where

$CH_{3}COOH(aq)$ act as Acid

$NaHCO_{3}(aq)$ act as weak base

Reactant : $CH_{3}COOH(aq)$ , $NaHCO_{3}$

Number of atoms of :

C = 2 ( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 2 + 1

= 3

H = 4( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 4 + 1

O = 2( $CH_{3}COOH(aq)$ ) + 3 ( $NaHCO_{3}$ )

= 5

Na = 1 ( $NaHCO_{3}$ )

= 1

Product : $CO_{2}(g)$ , $H_{2}O(l)$ , $CH_{3}COONa(aq)$

Number of atoms :

C = 1( $CO_{2}(g)$ ) + 2( $CH_{3}COONa(aq)$ )

= 1 + 2

= 3

H = 2( $H_{2}O(l)$ ) + 3( $CH_{3}COONa(aq)$ )

= 2 + 3

= 5

O = 1( $H_{2}O(l)$ ) + 2( $CH_{3}COONa(aq)$ )

+2( $CO_{2}(g)$

= 1 + 2 + 2

= 5

Na = 1( $CH_{3}COONa(aq)$

= 1

Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction

2. $CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + CO_{2}(g)$

This is double displacement reaction .

Check the balancing in both reactant and products should be :

Na = 2

H = 2

Ca = 1

C = 2

O = 6

Cl = 2

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Perchloric acid (HCO₄) is one of strongest inorganic acids. Perchloric acid precipitation is used to removes most of the protein present in the sample and stabilize many of the small molecule analytes. It can use also to precipitation glycogen, ATP, glutathione, antioxidants.
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2 years ago

Which Of the following compounds is an electrolyte?

Anettt [7]

Hello,

Here is your answer:

The proper answer to this question is option D "<span>sodium hydroxide".

Here is how:

Sodium Hydroxide its a white substance that is a </span><span>electrolyte.

Your answer is D.

If you need anymore help feel free to ask me!

Hope this helps!</span>

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3 years ago

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Softa [21]

Answer:

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c = fλ

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Explanation:

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