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3 years ago

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How many grams of CaCo3 are required to prepare 50.0g of CaO

1 answer:

olchik [2.2K]3 years ago

7 0

Answer is: 89 g of CaCO₃ is <span>equired to prepare 50.0g of CaO.
</span>Chemical reaction: CaCO₃ → CaO + CO₂.
m(CaO) = 50,0 g.
m(CaCO₃) = ?
n(CaO) = m(CaO) ÷ M(CaO).
n(CaO) = 50,0 g ÷ 56 g/mol.
n(CaO) = 0,89 mol.
from reaction: n(CaCO₃) : n(CaO) = 1 : 1.
n(CaCO₃) = n(CaO).
n(CaCO₃) = 0,89 mol.
m(CaCO₃) = m(CaCO₃) · M(CaCO₃).
m(CaCO₃) = 0,89 mol · 100 g/mol
m(CaCO₃) = 89 g.
n - amount of substance.

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<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

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where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

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We know that:

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<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

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In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

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To formulate the empirical formula, we need to follow some steps:

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<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

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