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solong [7]
3 years ago
5

Tin metal reacts with hydrogen fluoride to produce tin (ll )fluoride and hydrogen gas according to the balanced equation. Sn(s)+

2 HF(g) - SnF2(s)+H2(g)
Chemistry
1 answer:
ladessa [460]3 years ago
8 0

Answer:

                      1.263 moles of HF

Explanation:

                     The balance chemical equation for given single replacement reaction is;

                                    Sn + 2 HF → SnF₂ + H₂

Step 1: <u>Calculate Moles of Tin as;</u>

As we know,

                    Moles  =  Mass / A.Mass    ----- (1)

Where;

           Mass of Tin  =  75.0 g

           A.Mass of Tin  =  118.71 g/mol

Putting values in eq. 1;

                    Moles  =  75.0 g / 118.71 g/mol

                    Moles  =  0.6318 moles of Sn

Step 2: <u>Find out moles of Hydrogen Fluoride as;</u>

According to balance chemical equation,

                     1 mole of Sn reacted with  =  2 moles of HF

So,

              0.6318 moles of Sn will react with  =  X moles of HF

Solving for X,

                     X =  0.6318 mol × 2 mol / 1 mol

                     X  =  1.263 moles of HF

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Answer:

a. The limiting reactant is Ca(OH)₂

b. The theoretical yield of CaCl₂ is approximately 621.488 grams

c. The percentage yield of CaCl₂ is approximately 47.06%

Explanation:

a. The given chemical reaction is presented as follows;

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Therefore;

One mole of Ca(OH)₂ reacts with two moles of HCl to produce one mole of CaCl₂ and two moles of water H₂O

The mass of HCl in an experiment, m₁ = 229.70 g

The mass of Ca(OH)₂ in an experiment, m₂ = 207.48 g

The molar mass of HCl, MM₁ = 36.458 g/mol

The molar mass of Ca(OH)₂, MM₂ =74.093 g/mol

The number of moles of HCl, present, n₁ = m₁/MM₁

∴ n₁ = m₁/MM₁ = 229.70 g/(36.458 g/mol) ≈ 6.3 moles

The number of moles of Ca(OH)₂, present, n₂ = m₂/MM₂

∴ n₂ = m₂/MM₂ = 207.48 g/(74.093 g/mol) ≈ 2.8 moles

The number of moles of Ca(OH)₂, present, n₂ = 2.8 moles

According the chemical reaction equation the number of moles of HCl the 2.8 moles of Ca(OH)₂ will react with, = 2.8 × 2 moles = 5.6 moles of HCl

Therefore, there is an excess HCl in the reaction and Ca(OH)₂ is the limiting reactant

b. According the chemical reaction equation the number of moles of CaCl₂ produced in he reaction by the 2.8 moles of Ca(OH)₂ = 2.8 × 2 moles = 5.6 moles of CaCl₂

The molar mass of CaCl₂ = 110.98 g/mol

The mass of the 5.6 moles of CaCl₂ = 5.6 moles × 110.98 g/mol ≈ 621.488 grams

The theoretical yield of CaCl₂ ≈ 621.488 grams

c. Given that the actual mass of CaCl₂ produced = 292.5 grams, we have;

The percentage yield of CaCl₂ = The actual yield/(The theoretical yield) × 100

∴ The percentage yield of CaCl₂ = (292.5 g)/(621.488 g) × 100 ≈ 47.0644646397%

The percentage yield of CaCl₂ ≈ 47.06%.

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