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3 years ago

11

5. Hydrogen & oxygen react chemically to form water. How much water

1 answer:

stepladder [879]3 years ago

6 0

39.25 g of water (H₂O)

Explanation:

We have the following chemical reaction:

2 H₂ + O₂ → 2 H₂O

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of H₂ = 14.8 / 2 = 7.4 moles

number of moles of O₂ = 34.8 / 32 = 1.09 moles

We see from the chemical reaction that 2 moles of H₂ will react with 1 mole of O₂ so 7.4 moles of H₂ will react with 3.7 moles of O₂ but we only have 1.09 moles of O₂ available. The O₂ will be the limiting reactant. Knowing this we devise the following reasoning:

if 1 moles of O₂ produces 2 moles of H₂O

then 1.09 moles of O₂ produces X moles of H₂O

X = (1.09 × 2) / 1 = 2.18 moles of H₂O

mass = number of moles × molar weight

mass of H₂O = 2.18 × 18 = 39.25 g

Learn more about:

limiting reactant

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4. Al2O3 (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H20(1) Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.

slava [35]

Answer:

Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over

Explanation:

For the reaction:

Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)

10,0g of Al₂O₃ are:

10,0g ₓ $\frac{1mol}{102g}$ = <em>0,0980 moles</em>

And 10,0g of HCl are:

10,0 gₓ $\frac{1mol}{36,5g}$ = <em>0,274 moles</em>

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For a total reaction of 0,274 moles of HCl you need:

0,274× $\frac{1molesAl_{2}O_3}{6 mole HCl}$ = <em>0,0457 moles of Al₂O₃</em>

Thus, limiting reactant is HCl

The grams produced of AlCl₃ are:

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<em />

The moles of Al₂O₃ that don't react are:

0,0980 moles - 0,0457 moles =<em> </em>0,0523 moles

And its mass is:

0,0523 molesₓ $\frac{102g}{1mol}$ = <em>5,33 g of Al₂O₃ </em>

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I hope it helps!

7 0

3 years ago