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3 years ago

35g of HF was prepared by reacting 112g of CaF2 with an excess of H2SO4 calculate the percentage yield

Chemistry

1 answer:

Tom [10]3 years ago

7 0

Answer: 61%

The reaction equation should be

CaF2 + H2SO4 → 2HF + CaSO4

For every 1 molecule CaF2 used, there will be 2 molecules of HF formed. The molecular mass of CaF2 is 78/mol while the molecular mass of HF is 20g/mol. If the yield is 100%, the amount of HF formed by 112g CaF2 would be: 112g/(78g/mol) * 2 * (20g/mol)=57.43g

The percentage yield of the reaction would be: 35g/57.43g= 60.94%

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A chemist measures the enthalpy change ?H during the following reaction: Fe (s) + 2HCl (g) ? FeCl2 (s) + H2 (g) =?H?157.kJ Use t

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Correct Question:

A chemist measures the enthalpy change ΔH during the following reaction: Fe(s) + 2HCl(g)-->FeCl2(s) + H2 ΔH=-157.0 kJ. Use this information to complete the table below. Round each of your answers to the nearest kJ/mol

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+157 kJ

Explanation:

The enthalpy change of a reaction measures the amount of heat that is lost or gained by it. If ΔH >0 the heat is gained, and the reaction is called endothermic, if ΔH<0, the heat is lost, and the reaction is called exothermic.

If the reaction is inverted, the value of ΔH is inverted too (the opposite endothermic reaction is exothermic), and if the reaction is multiplied by a constant, ΔH will be multiplied by it too.

1) 2Fe(s) + 4HCl --> 2FeCl2(s) + 2H2(g)

This reaction is the product of the given reaction by 2, so

ΔH = 2*(-157) = -314 kJ

2) 4FeCl2(s) + 4H2(g) --> 4Fe(s) + 8HCl(g)

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Answer:

option a) 3 g

Explanation:

mass of Glucose = 5 g

Mass of H₂O = ?

Reaction Given:

6CO₂ + 6H₂O ----> C₆H₁₂O₆ + 6O₂

Solution:

First we have to find mass of glucose from balanced reaction.

So,

Look at the reaction

6CO₂ + 6H₂O -------> C₆H₁₂O₆ + 6O₂

6 mol 1 mol

As 6 mole of water (H₂O) give 1 mole of Glucose (C₆H₁₂O₆ )

Convert moles to mass

molar mass of C₆H₁₂O₆ = 6(12) + 12(1) + 6(16)

molar mass of C₆H₁₂O₆ = 72 + 12 + 96

molar mass of C₆H₁₂O₆= 180 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now

6CO₂ + 6H₂O ---------> C₆H₁₂O₆ + 6O₂

6 mol (18 g/mol) 1 mol (180 g/mol)

108 g 180 g

108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆)

if 108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆) so how many grams of water (H₂O) will be required to produce 5 g of glucose (C₆H₁₂O₆).

Apply Unity Formula

108 g of water (H₂O) ≅ 180 g of glucose (C₆H₁₂O₆)

X g of water (H₂O) ≅ 5 g of glucose (C₆H₁₂O₆)

Do cross multiply

mass of water (H₂O) = 108 g x 5 g / 180 g

mass of water (H₂O) = 3 g

So 3 g of water is required to produce 5 g of glucose.

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