How much co2 is realsed by burning trash

Answer is: volume of H₂SO₄ is 42.1 mL.
Chemical reaction: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
c(H₂SO₄) = 0,4567 M = 0,4567 mol/L.
V(NaOH) = 30 mL ÷ 1000 mL/L = 0,03 L.
c(NaOH) = 0,321 M = 0,321 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,321 mol/L · 0,030 L.
n(NaOH) = 0,00963 mol.
From chemical reaction: n(H₂SO₄) : n(NaOH) = 1 : 2.
n(H₂SO₄) = 0,01926 mol.
V(H₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).
V(H₂SO₄) = 0,01926 mol ÷ 0,4567 mol/L.
V(H₂SO₄) = 0,0421 L = 42,1 mL.

3 0

3 years ago

Given that a 11.00 g milk bar chocolate bar contains 7.500 g of sugar, calculate the percentage of sugar present in 11.00 g of m

alukav5142 [94]

Answer:

68.18%

Explanation:

The question asking for the percentage of sugar present in milk bar chocolate. To find the percentage of a certain molecule, you have to divide the mass of the molecule with the total mass of the product. The calculation will be:

sugar percentage = sugar weight / milk bar chocolate weight * 100%

sugar percentage = 7.5g / 11g * 100%= 68.18%

7 0

3 years ago

A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.25 grams of aluminum foil in a solut

mrs_skeptik [129]

Answer: The correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.

Explanation:

To calculate the number of moles, we use the formula:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of Aluminium:

Molar mass of aluminium = 27 g/mol

Given mass of aluminium = 0.25g

Putting values in above equation, we get:

$Moles=\frac{0.25g}{27g/mol}=0.00925mol$

Moles of Copper chloride:

Molar mass of copper chloride = 134.45 g/mol

Given mass of copper chloride= 0.40g

Putting values in above equation, we get:

$Moles=\frac{0.40g}{134.45g/mol}=0.00297mol$

For the given chemical equation:

$4Al(s)+3CuCl_2(aq.)\rightarrow 2Al_2Cl_3(aq.)++3Cu(s)$

By Stoichiometry,

3 moles of Copper chloride reacts with 4 moles of aluminium

So, 0.00297 moles of copper chloride will react with = $\frac{4}{3}\times 0.00297$ = 0.00396 moles of Aluminium.

As, the required moles of aluminium is less than the given moles of aluminium. Hence, it is considered as the excess reagent.

Copper chloride is the limiting reagent in the given chemical reaction because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of copper chloride produces 3 moles of copper metal.

So, 0.00297 moles of copper chloride will produce = $\frac{3}{3}\times 0.00297$ = 0.00297 moles of copper metal.

Now, to calculate the given mass of copper metal, we use the equation required to calculate moles:

Molar mass of copper = 63.5 g/mol

Putting values in that equation, we get:

$0.00297mol=\frac{\text{given mass}}{63.5g/mol}$

Given mas of copper metal = 0.188grams (approx 0.20grams)

For the given reaction, some of the aluminum is left behind because it is the excess reagent in the reaction.

Hence, the correct answer is about 0.20 grams of copper (II) is formed, and some aluminum is left in the reaction mixture.

6 0

3 years ago

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