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3 years ago

How many grams of H2O are produced when 35.0 g of NaOH reacts with 17.5 g of CO,?

Chemistry

1 answer:

zhenek [66]3 years ago

6 0

Answer:

2NaOH + CO2 -> Na2CO3 + H2O

1) Find the moles of each substance

$\eq n(NaOH)=\frac{35.0}{22.99+16.00+1.008\\}\ =\frac{35.0}{39.998} \ = 0.8750437522 moles\\n(CO_{2} ) = \frac{17.5}{12.01+32.00} = \frac{17.5}{44.01} = 0.3976369007 moles\\$

2) Determine the limitting reagent

$\\NaOH = \frac{0.8750437522}{2} = 0.4375218761\\\\$

∴ Carbon dioxide is limitting as it has a smaller value.

3) multiply the limiting reagent by the mole ratio of unknown over known

n(H2O ) = 0.3976369007 × 1/2

= 0.1988184504 moles

4) Multiply the number of moles by the molar mass of the substance.

m = 0.1988184504 × (1.008 × 2 + 16.00)

= 0.1988184504 × 18.016

= 3.581913202 g

Explanation:

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Answer:

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Let the age of the sample be t years old.

Initial mass percentage of carbon-14 in an artifact = 100%

Initial mass of carbon-14 in an artifact = $[A_o]$

Final mass percentage of carbon-14 in an artifact t years = 60%

Final mass of carbon-14 in an artifact = $[A]=0.06[A_o]$

Half life of the carbon-14 = $t_{1/2}=5730 years$

$k=\frac{0.693}{t_{1/2}}$

$[A]=[A_o]\times e^{-kt}$

$[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}$

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Solving for t:

t = 4223.71 years ≈ 4224 years

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