Which increases the rate of a chemical reaction?

If 125 g of CaCO3 is mixed and reacted with 125 g of HCl, which reactant is limiting and how many grams of CO2 can be made?

Trava [24]

Answer:

CaCO3 is the limiting reactant

55 g of CO2 is made

Explanation:

First we must put down the reaction equation;

CaCO3(s) + 2HCl(aq) ---------> CaCl2(s) + H2O(l) + CO2(g)

Number of mole of CaCO3 = 125g/100gmol-1 = 1.25 moles

From the reaction equation;

1 mole of CaCO3 yields 1 mole of CO2

Hence 1.25 moles of CaCO3 yields 1.25 moles of CO2

For HCl;

number of moles of HCl = 125g/36.5 g mol-1 = 3.42 moles

From the reaction equation;

2 moles of HCl yields 1 mole of CO2

3.42 moles of HCl yields 3.42 * 1/2 = 1.71 moles of CO2

Hence CaCO3 is the limiting reactant.

Mass of CO2 produced = 1.25g * 44 gmol-1 = 55 g of CO2

6 0

2 years ago

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for

tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. $K(s)+O_2(g)\rightarrow KO_2(s)$

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$

Answer: A. $K(s)+O_2(g)\rightarrow KO_2(s)$ : decreases

B. $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2(g)$ : decreases

C. $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ : no change

D. $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ : increases

Explanation:

Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction $K(s)+O_2(g)\rightarrow KO_2(s)$ , as gaseous reactant is changed to solid product, entropy decreases.

In reaction $CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)$ , as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.

In reaction $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ , as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.

In reaction $N_2O_2(g)\rightarrow 2NO(g)+O_2(g)$ , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

7 0

3 years ago

Does sugar can dissolve in ccl4

wel

Answer:

Sugar is a polar compound , in the solid state of it Hydrogen bond strongly binds the molicules of sugar together. Being a non polar solvent and low polarity difference between Cl and C atom C—Cl bond in CCl4 is unable to form hydrogen bond. That's why sugar can't be soluble in CCl4.

Explanation:

8 0

2 years ago

Determine the molar concentration of na+ and po4 3- in a 2.25 M Na3 PO4 solution

muminat

Answer:

A. The concentration of Na^+ in the solution is 6.75 M.

B. The concentration of PO4^3- in the solution is 2.25 M.

Explanation:

We'll begin by writing the balanced dissociation equation for Na3PO4.

This is illustrated below:

Na3PO4 will dissociate in solution as follow:

Na3PO4(aq) —> 3Na^+(aq) + PO4^3-(aq)

Thus, from the balanced equation above,

1 mole of Na3PO4 produce 3 moles of Na^+ and 1 mole of PO4^3-

A. Determination of the concentration of Na+ in 2.25 M Na3PO4 solution.

This can be obtained as follow:

From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of Na^+.

Therefore, 2.25 M Na3PO4 solution will produce = (2.25 x 3) /1 = 6.75 M Na^+.

Therefore, the concentration of Na^+ in the solution is 6.75 M

B. Determination of the concentration of PO4^3- in 2.25 M Na3PO4 solution.

This can be obtained as follow:

From the balanced equation above,

1 mole of Na3PO4 produce 1 mole of PO4^3-

Therefore, 2.25 M Na3PO4 solution will also produce 2.25 M PO4^3-.

Therefore, the concentration of PO4^3- in the solution is 2.25 M.

7 0

3 years ago

1) A common experiment to determine the relative reactivity of metallic elements is to place a pure sample of one metal into an

Annette [7]

Answer: $Zn(s)+CuSO_4(aq)\rightarrow ZnSO_4+Cu$

Explanation:-

Single replacement reaction is a chemical reaction in which more reactive element displaces the less reactive element from its salt solution.

As zinc is more reactive than copper, it could easily displace copper from its aqueous solution and thus leads to formation of zinc (II) sulfate and pure copper.

The chemical reaction can be represented as :

$Zn(s)+CuSO_4(aq)\rightarrow ZnSO_4+Cu$

The phases are represented as (s) for solid sate, (l) for liquid state, (g) for gaseous state and (aq) for aqueous state.

7 0

3 years ago

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