Question

Heyy guys, so basically i need help with stoichiometric calculation I will give you 100 points just to answer all of these answers accurately with working out (ps ill mark you brainliest x), thanks. PLEASE HELPPP. I'm desperate x
3. What masses of ethanol and ethanoic acid would need to be reacted together to give 1 g of ethyl ethanoate?

C^2H^5OH + CH^3CO^2 H → CH^3CO^2C^2H^5 + H^2O


4. What mass of iron(III) oxide would need to be reduced to produce 100 tonnes of iron in a blast furnace?

Fe^2^O^3 + CO → Fe + CO^2


5. What mass of silver nitrate as a solution in water would need to be added to 5 g of sodium chloride to
ensure complete precipitation of the chloride?

AgNO^3(aq) + NaCl (aq) → AgCl (s) + NaNO^3(aq)

6. Copper(II) oxide reacts with sulphuric acid to produce copper(II) sulphate. If this is allowed to crystallise the formula of the crystals is CuSO 4 .5H 2 O. What mass of copper oxide would be needed to produce 100 g of crystals?

CuO + H^2O + H^2SO^4 = CuSO^4 .5H^2O


7. In the following reactions calculate the mass of precipitate formed from 20 g of the metal salt in each case.
a. ZnSO^4 (aq) + 2NaOH → Zn(OH)^2(s) + Na^2SO^4(aq)
b. Al^2 (SO^4 ) 3(aq) + 6NaOH → 2Al(OH)^3(s) + 3Na^2SO^4(aq)
c. MgSO^4(aq) + 2NaOH → Mg(OH)^2(s) + Na^2^SO^4(aq)

Answer 1

Answer:

Explanation:

i will show in details how 2 do the 1st Q n u can do the rest by following the way how it is done

3. given C^2H^5OH + CH^3CO^2H → CH^3CO^2C^2H^5 + H^2O

molar ratio of ethanol, ethanoic acid and ethyl ethanoate is 1 : 1 : 1

so mass = no. of moles * molecular mass

for same no. of moles, mass / molecular mass is the same

molecular mass of CH^3CO^2C^2H^5 = 12+1*3+12+16*2+12*2+5*1 = 88

molecular mass of C^2H^5OH = 12*2+1*5+16+1 = 46

molecular mass of CH^3CO^2H = 12+1*3+12+16*2+1 = 60

1 g of ethyl ethanoate = 1/88 mole

it requires 1/88*46 = 0.5227 g of ethanol; and

1/88*60 = 0.6818 g of ethanoic acid

to react together to form 1 g of ethyl ethanoate

Answer 2

Answer:

3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

5. The mass of silver nitrate required is approximately 14.53 grams

6. The mass of copper oxide that would be needed is approximately 31.86 grams

7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams

b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams

c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams

Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g

The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g

4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles

The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes

5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles

The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams

6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles

The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams

7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles

2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams

b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams