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zepelin [54]
2 years ago
5

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]2 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:

Here's what I get  

Explanation:

1. Balanced equation

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2. Mechanism

HQ⁻ + CH₃-Br ⟶[HQ···CH₃···Br]⁻⟶ HQ-CH₃ + Br⁻

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The hydroxide ion attacks the back side of the carbon atom in the bromomethane (A).

At the same time as the Q-H bond starts to form, the C-Br bond starts to break.

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8 0
3 years ago
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RFM 24 RFM 95

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