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2 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]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:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

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$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

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The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

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So, the percentage component of calcium is:

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$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

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The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

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So, the percentage component of oxygen is:

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