CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+ CO2(g) CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+ CO2(g) How many grams of calcium chloride will be produced when 28.0 g28.0 g of calcium carbonate is combined with 11.0 g11.0 g of hydrochloric acid? mass of CaCl2CaCl2: gg Which reactant is in excess? CaCO3CaCO3 HClHCl How many grams of the excess reactant will remain after the reaction is complete?

A. 16.73g of CaCl2 is produced

B. CaCO3

C. Excess mass of CaCO3 is 12.93g

Explanation:

The balanced equation for the reaction.

CaCO3(s) + 2HCl(aq) —> CaCl2(aq)

+ H2O(l) + CO2(g)

A. Step 1:

Determination of the masses of CaCO3 and HCl that reacted from the balanced equation. This is illustrated below:

CaCO3(s) + 2HCl(aq) —> CaCl2(aq)

+ H2O(l) + CO2(g)

Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol

Molar Mass of HCl = 1 + 35.5 = 36.5g/mol

Mass of HCl from the balanced equation = 2 x 36.5 = 73g

From the balanced equation above,

100g of CaCO3 reacted.

73g of HCl reacted.

A. Step 2:

Determination of the limiting reactant.

Let consider using all the 28g of CaCO3 to see if there will be any leftover for HCl.

From the balanced equation above,

100g of CaCO3 reacted with 73g of HCl.

Therefore, 28g of CaCO3 will react with = (28 x 73)/100 = 20.44g of HCl.

The mass of HCl obtained from the above calculation is far above the mass (11g) that was given in the question.

Now let us consider using all the 11g of HCl given from the question to see if there will be any leftover for CaCO3. This is illustrated below:

From the balanced equation above,

100g of CaCO3 reacted with 73g of HCl.

Therefore, Xg of CaCO3 will react with 11g of HCl i.e

Xg of CaCO3 = (11 x100) /73

Xg of CaCO3 = 15.07g

Now we can see clearly that there are leftover for CaCO3 as only 15.07g reacted out of 28g that was given.

Therefore, HCl is the limiting reactant and CaCO3 is the excess reactant.

A. Step 3:

Determination of the mass of CaCl2 produced from the balanced equation.

CaCO3(s) + 2HCl(aq) —> CaCl2(aq)

+ H2O(l) + CO2(g)

Molar Mass of CaCl2 = 40 + (35.5x2) = 40 + 71 = 111g/mol

From the balanced equation above,

111g of CaCl2 is produced.

A. Step 4:

Determination of the mass of CaCl2 that will be produced when 28g of CaCO3 combined with 11g of HCl.

CaCO3(s) + 2HCl(aq) —> CaCl2(aq)

+ H2O(l) + CO2(g)

From the balanced equation above,

73g of HCl produced 111g of CaCl2.

Therefore, 11g of HCl will produce = (11 x 111)/73 = 16.73g of CaCl2

B. From the calculations made above, CaCO3 is the excess reactant.

C. Determination of the excess mass remaining after the reaction.

The ecx reactant is CaCO3.

Mass of CaCO3 given = 28g

Mass of CaCO3 that reacted = 15.07g

Excess mass of CaCO3 = (Mass of CaCO3 given) - (Mass of CaCO3 that reacted)

Excess mass of CaCO3 = 28 - 15.07

Excess mass of CaCO3 = 12.93g

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j