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Aleks [24]
3 years ago
14

How many moles of sand (SiO2) are in 30 g of sand?

Chemistry
2 answers:
luda_lava [24]3 years ago
6 0

Answer:

\boxed {\boxed {\sf About \ 0.5 \ moles \ of \ SiO_2}}

Explanation:

To convert from grams to moles, we must use the molar mass.

<u>1. Molar Mass </u>

Use the Periodic Table to find the masses of the individual elements (silicon and oxygen) in sand.

  • Silicon (Si): 28.085 g/mol
  • Oxygen (O): 15.999 g/mol

Examine the formula for sand: SiO₂. There is a subscript of 2 after oxygen, so there must be 2 oxygen atoms. Multiply oxygen's mass by 2 and add silicon's mass to find the molar mass of sand.

  • SiO₂: 2(15.999 g/mol) + 28.085 g/mol= g/mol

<u>2. Calculate Moles </u>

Use the molar mass as ratio.

\frac{60.083 \ g \ SiO_2}{1 \ mol \ SiO_2}

Multiply by the given number of grams (30)

30 \ g \ SiO_2 *   \frac{60.083 \ g \ SiO_2}{1 \ mol \ SiO_2}

Flip the fraction so the grams of sand will cancel.

30 \ g \ SiO_2 *\frac{1 \ mol \ SiO_2}{60.083 \ g \ SiO_2}

30  *\frac{1 \ mol \ SiO_2}{60.083 }

\frac{30 \ mol \ SiO_2}{60.083 }

0.499309289 \ mol \ SiO_2

<u>3. Round </u>

The original measurement of grams has 1 signfiicant figure. We must round our answer to 1 sig fig.

For the answer we found, that is the tenth place. The 9 in the hundredth tells us to round the 4 to a 5.

\approx 0.5 \ mol \ SiO_2

There are about <u>0.5 moles</u> of SiO₂ in 30 grams.

anastassius [24]3 years ago
5 0
<h3>Answer:</h3>

0.50 mol SiO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

30 g SiO₂ (sand)

<u>Step 2: Identify Conversions</u>

Molar Mass of Si - 28.09 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of SiO₂ - 28.09 + 2(16.00) = 60.09 g/mol

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 30 \ g \ SiO_2(\frac{1 \ mol \ SiO_2}{60.09 \ g \ SiO_2})
  2. Multiply/Divide:                  \displaystyle 0.499251 \ mol \ SiO_2

<u>Step 4: Check</u>

<em>Follow sig figs and round. We are given 2 sig figs.</em>

0.499251 mol SiO₂ ≈ 0.50 mol SiO₂

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Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
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Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

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