An organic liquid is a mixture of methyl alcohol (ch3oh m ch_3oh) and ethyl alcohol (c2h5oh m c_2h_5oh ). a 0.220-g m g sample of the liquid is burned in an excess of o2(g) m o_2(g) and yields 0.367g g co2(g) m co_2(g) (carbon dioxide). set up two algebraic equations, one expressing the mass of carbon dioxide produced in terms of each reagent and the other expressing the mass of sample burned in terms of each reagent.

what is the mass of methyl alcohol (ch3oh m ch_3oh) in the sample?

The mass of methanol in the sample is 0.1002 grams

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

For carbon dioxide:

Given mass of carbon dioxide = 0.367 g

Molar mass of carbon dioxide = 44.0 g/mol

Putting values in equation 1, we get:

Let the mass of methanol in the sample be 'x' grams.

We are given:

Mass of the mixture of methanol and ethanol = 0.220 g

Mass of methanol = x g

Mass of ethanol = (0.220 - x) g

For methanol:

Given mass of methanol = x g

Molar mass of methanol = 32 g/mol

Putting values in equation 1, we get:

For ethanol:

Given mass of ethanol = (0.220 - x) g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

The chemical equation for the combustion of methanol follows:

By Stoichiometry of the reaction:

1 mole of methanol produces 1 mole of carbon dioxide

So, moles of carbon dioxide

The chemical equation for the combustion of ethanol follows:

By Stoichiometry of the reaction:

1 mole of ethanol produces 2 moles of carbon dioxide

So, moles of ethanol will produce = moles of carbon dioxide

Total moles of carbon dioxide:

Hence, the mass of methanol in the sample is 0.1002 grams

4- c, because all matter stays the same no matter what. it is still there. 5- b, because matter is known as mass. 6-c because, flammability is a chemical property. malleability is changing the shape of an object.

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