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2 years ago

In the combustion analysis of 0.1127 g of glucose (c6h12o6), what mass, in grams, of co2 would be produced?

Chemistry

1 answer:

Charra [1.4K]2 years ago

8 0

The combustion produces 0.1652 g CO₂.

Balanced equation = C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

Moles of C₆H₁₂O₆ = 0.1127 g C₆H₁₂O₆ × (1 mol C₆H₁₂O₆/180.16 g C₆H₁₂O₆)

= 0.000 6256 mol C₆H₁₂O₆

Moles of CO₂ = 0.000 6256 mol C₆H₁₂O₆× (6 mol CO₂/1 mol C₆H₁₂O₆)

= 0.003 753 mol CO₂

Mass of CO₂ = 0.003 753 mol CO₂ × (44.01g CO₂/1 mol CO₂) = 0.1652 g CO₂

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2 years ago

If 40.0 mL of a calcium nitrate solution reacts with excess potassium carbonate to yield 0.524 grams of a precipitate, what is t

tankabanditka [31]

Answer : The molarity of calcium ion on the original solution is, 0.131 M

Explanation :

The balanced chemical reaction is:

$Ca(NO_3)_2+K_2CO_3\rightarrow CaCO_3+3KNO_3$

When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.

First we have to calculate the moles of $CaCO_3$

$\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}$

Given:

Mass of $CaCO_3$ = 0.524 g

Molar mass of $CaCO_3$ = 100 g/mol

$\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol$

Now we have to calculate the concentration of $CaCO_3$

$\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M$

Now we have to calculate the concentration of calcium ion.

As, calcium carbonate dissociate to give calcium ion and carbonate ion.

$CaCO_3\rightarrow Ca^{2+}+CO_3^{2-}$

So,

Concentration of calcium ion = Concentration of $CaCO_3$ = 0.131 M

Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M

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3 years ago

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Answer:

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