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RSB [31]
3 years ago
11

Balance the following Equation:

Chemistry
1 answer:
lisabon 2012 [21]3 years ago
7 0

Answer:

HCl is the limiting reactant. It will completely be consumed (1.37 moles)

Option D is correct

Explanation:

Step 1: Data given

Mass of Zinc (Zn) = 50.0 grams

Mass of Hydrogen chloride (HCl) = 50.0 grams

atomic mass Zn = 65.38 g/mol

Molar mass HCl = 36.46 g/mol

Step 2: The balanced equation

Zn + 2HCl → ZnCl2 + H2

Step 3: Calculate moles

Moles = mass / molar mass

Moles Zn = 50.0 grams / 65.38 g/mol

Moles Zn = 0.764 moles

Moles HCl = 50.0 grams / 36.46 g/mol

Moles HCl = 1.37 moles

Step 4: Calculate limiting reactant

For 1 mol Zn we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2

HCl is the limiting reactant. It will completely be consumed (1.37 moles)

Zn is in excess. There will react 1.37/2 = 0.685 moles

There will remain 0.764 -0.685 = 0.079 moles

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opper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f =
Y_Kistochka [10]

Answer:

55.373g/l

Explanation:

The dissolved amount of sparingly soluble salts is interlinked with a unitlesss quantity called as solubility product. It is a fixed quantity that only increases with the rise in temperatures and is used to predict the salting out of compounds. If the value of ionic product (Q) is larger than (Ksp), precipitation of compound occurs.

Given:

The solubility product of CuBr is 6.3×10−9.

The concentration of NH3 is 0.10 M.

Formula and Calculations:

The dissolution reaction (I) of CuBr is shown below.

The reaction showing dissolution of CuBr in NH3 is shown below.

The above reaction can be obtained by adding reaction (I) and (II) as shown below.

The equilibrium constants will get multiplied.

Suppose the solubility of CuBr is “s”.

It is given that concentration of NH3 is 0.10 M.

The equilibrium constant expression for the above reaction is as follows,

Here,

The concentration of pure solids is 1 M. Thus, the concentration of CuBr is 1 M.

As calculated, the value of Ksp is 396.9.

Substitute all the required values in above formula.  

On further solving above equation,

Therefore, the solubility of CuBr in ammonia is 0.386 M.

The formula to calculate solubility

Solubuility (g/l)= Molarity(M) x Molarmass

Chemistry homework question answer, step 2, image 10

The molar mass of CuBr is 143.45 g/mol.

The formula to calculate solubility in g/L is given below.

The molar mass of CuBr is 143.45 g/mol.

therefore,

solubility = 0.386M x 143.45g/mol

where (M = mol/l)

solubility = 55.375g/l

5 0
3 years ago
How can the law of conservation of<br> mass for a chemical reaction be expressed?
djverab [1.8K]

Answer:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

Explanation:

How does the concept of conservation of mass apply to chemical reactions? the reactants and products have exactly the same atoms. the reactants and products have exactly the same molecules. the change in the amount of matter is equal to the change in energy.

can someone help me with my qustions?

5 0
3 years ago
Calculate the solubility (in M units) of ammonia gas in water at 298 K and a partial pressure of 7.00 bar . The Henry’s law cons
Juliette [100K]

Explanation:

just for friendship kvp-hxns-jst

6 0
3 years ago
. The dissolving process is exothermic when the energy
Serhud [2]

Answer:

A

Explanation:

The dissolving process depends on the interaction between solute and solvent (solvation) and the breaking up of the intermolecular bond between solutes. The former is exothermic in nature, while the later is endothermic. Energy is released when solute-solvent particles interact. When this energy exceeds the energy required to break intermolecular bonds between the solute particles, dissolution is exothermic.

8 0
3 years ago
How many grams of oxygen (O) are present in 0.0207 moles of Ca(HCO 3) 2
Zolol [24]

Answer:

1.99grams

Explanation:

- First, we need to calculate the molar mass of the compound: Ca(HCO3)2

Ca = 40g/mol, H = 1g/mol, C = 12g/mol, O = 16g/mol

Hence, Ca(HCO3)2

= 40 + {1 + 12 + 16(3)}2

= 40 + {13 + 48}2

= 40 + {61}2

= 40 + 122

= 162g/mol

Molar mass of Ca(HCO3)2 = 162g/mol

- Next, we calculate the mass of oxygen in one mole of the compound, Ca(HCO3)2.

Oxygen = {16(3)}2

= 48 × 2

= 96g of Oxygen

- Next, we calculate the percentage composition of oxygen by mass by dividing the mass of oxygen in the compound by the molar mass of the compound i.e.

% composition of O = 96/162 × 100

= 0.5926 × 100

= 59.26%.

- The number of moles of the compound, Ca(HCO3)2, must be converted to mass by using the formula; mole = mass/molar mass

0.0207 = mass/162

Mass = 162 × 0.0207

Mass = 3.353grams

However, in every gram of Ca(HCO3)2, there is 0.5926 g of oxygen

Hence, in 3.353grams of Ca(HCO3)2, there will be;

0.5926 × 3.353

= 1.986

= 1.99grams.

Therefore, there is 1.99grams of Oxygen in 0.0207 moles (3.353g) of Ca(HCO3)2.

5 0
3 years ago
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