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Ierofanga [76]
3 years ago
13

How many grams of a 14.0% (w/w) sugar solution contain 62.5 of sugar?

Chemistry
1 answer:
Delicious77 [7]3 years ago
3 0

Answer:

m_{solution}=446.4g

Explanation:

Hello,

In this case, by-mass percent is computed in terms of the mass of the solute and the mass of the solvent as shown below:

\% m=\frac{m_{solute}}{m_{solution}} *100\%

Thus, solving for the mass of the solution, we obtain:

\frac{x}{y} m_{solution}=\frac{m_{solute}}{\% m}*100\%\\ \\m_{solution}=\frac{62.5g}{14.0\%}*100\% \\\\m_{solution}=446.4g

Regards.

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The balanced chemical reaction is expressed as:

<span>8SO2 + 16H2S = 3S8 + 16H2O

We are given the initial amount of the reactants. From there, we determine the limiting reactant. We do as follows:

87.0 g SO2 ( 1 mol / 64.07 g ) = 1.36 mol SO2 ( 16 mol H2S / 8 mol SO2 ) = 2.72 mol H2S
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Therefore, the limiting reactant would be H2S. We calculate the maximum amount of S8 that can be produced from the amount of H2S.

2.55 mol H2S ( 3 mol S8 / 16 mol H2S ) ( 256.48 g / 1 mol ) = 122.63 g S8</span>
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A 1.0857 gram pure sample of a compound containing only carbon, hydrogen, and oxygen was burned in excess oxygen gas. 2.190 g of
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Answer:

  • C₂ H₄ O

Explanation:

<u>1) Mass of carbon (C) in 2.190 g of carbon dioxide (CO₂)</u>

  • atomic mass of C: 12.0107 g/mol
  • molar mass of CO₂: 44.01 g/mol
  • Set a proportion: 12.0107 g of C / 44.01 g of CO₂ = x / 2.190 g of CO₂
  • Solve for x:

         x = (12.0107 g of C / 44.01 g of CO₂ ) × 2.190 g of CO₂ = 0.59767 g of C

<u />

<u>2) Mass of hydrogen (H) in 0.930 g of water (H₂O)</u>

  • atomic mass of H: 1.00784 g/mol
  • molar mass of H₂O: 18.01528 g/mol
  • proportion: 2 × 1.00784 g of H / 18.01528 g of H₂O = x / 0.930 g of H₂O
  • Solve for x:

        x = ( 2 × 1.00784 g of H / 18.01528 g of H₂O) × 0.930 g of H₂O = 0.10406 g of H

<u>3) Mass of oxygen (O) in 1.0857 g of pure sample</u>

  • Mass of O = mass of pure sample - mass of C - mass of H
  • Mass of O = 1.0857 g - 0.59767 g - 0.10406 = 0.38397 g O

Round to four decimals: Mass of O = 0.3840 g

<u>4) Mole calculations</u>

Divide the mass in grams of each element by its atomic mass:

  • C: 0.59767 g / 12.0107 g/mol = 0.04976 mol
  • H: 0.10406 g / 1.00784 g/mol = 0.10325 mol
  • O: 0.3840 g / 15.999 g/mol = 0.02400 mol

<u>5) Divide every amount by the smallest value (to find the mole ratios)</u>

  • C: 0.04976 mol / 0.02400 mol = 2.07 ≈ 2
  • H: 0.10325 mol / 0.02400 mol = 4.3 ≈ 4
  • O: 0.02400 mol / 0.02400 mol = 1

Thus the mole ratio is 2 : 4 : 1, and the empirical formula is:

  • <u>C₂ H₄ O </u>← answer
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