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ehidna [41]
2 years ago
7

What mass of water (in grams) is produced by the reaction of 23.0 g of SiO2?

Chemistry
1 answer:
nikitadnepr [17]2 years ago
5 0

The mass of water produced by the reaction of the 23 g of SiO_2  is 13.8 g.

The given chemical reaction;

4Hf (g)  \ + \ SiO_2 (s) \ --> \ SiF_4(g) \ + \ 2H_2O(l)

In the given compound above, we can deduce the following;

  • molecular mass of SiO_2 = 28 + (2 x 16) = 60 g
  • molecular mass of 2H_2O = 2(18) = 36 g

60 g of SiO_2  --------- 36 g of water

23 g of SiO_2  ------------- ? of water

mass \ of \ water = \frac{23 \times 36}{60} = 13.8 \ g \ of \ water

Thus, the mass of water produced by the reaction of the 23 g of SiO_2  is 13.8 g.

  • <em>"Your question is not complete, it seems to be missing the following information";</em>

In the reaction of the given compound, 4Hf (g)  \ + \ SiO_2 (s) \ --> \ SiF_4(g) \ + \ 2H_2O(l), what mass of water (in grams) is produced by the reaction of 23.0 g of SiO2?

Learn more here:brainly.com/question/13644576

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The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) H
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Answer: The mass of O_2 needed is, 13.59 grams.

Explanation : Given,

Mass of C_3H_8 = 89.2 g

Molar mass of C_3H_8 = 42 g/mol

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First we have to calculate the moles of C_3H_8

\text{Moles of }C_3H_8=\frac{\text{Given mass }C_3H_8}{\text{Molar mass }C_3H_8}

\text{Moles of }C_3H_8=\frac{89.2g}{42g/mol}=2.124mol

Now we have to calculate the moles of C_3H_8

The balanced chemical equation is:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

From the reaction, we conclude that

As, 1 mole of C_3H_8 react with 5 mole of O_2

So, 2.124 mole of C_3H_8 react with \frac{2.124}{5}=0.4248 mole of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(0.4248moles)\times (32g/mole)=13.59g

Therefore, the mass of O_2 needed is, 13.59 grams.

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4 years ago
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