An organic liquid is a mixture of methyl alcohol (ch3oh) and ethyl alcohol (c2h5oh). a 0.220-g sample of the liquid is burned in an excess of o2(g) and yields 0.386 g co2(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) in the sample?

the mass of methyl alcohol (CH3OH) in the sample is 0.0641 g

Explanation:

1. Preliminars;

Mass  of CH₃ OH: MMass of C₂H₅OH: EMolar mass of CH₃ OH: 32.04 g/molMolar mass of C₂H₅OH: 46.07 g/molMolar mass of CO₂: 44.01 g/molMolar mass equation: molar mass = mass in grams / number of moles

2. Algebraic equation # 1: mass of sample burned in terms of each reagent.

M + E = 0.220 (1)

3. Algebraic equation # 2: mass of carbon dioxide produced in terms of each reagent.

a) Combustion of methyl alcohol

2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(g) 2 mol CH₃OH → 2 mol CO₂2 × 32.04 g CH₃OH → 2 × 44.01 g CO₂Mass of CO₂ produced from M grams of CH₃OH = M / 32.04 × 44.01 = 1.374 M

b) Combustion of ethyl alcohol

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g) 1 mol C₂H₅OH → 2 mol CO₂46.07 g CH₃OH → 2 × 44.01 g CO₂Mass of CO₂ produced from E grams of C₂H₅OH = E / 46.07 × 88.02 = 1.911 E

c) Mass of CO₂ (g) obtained, in terms of each reagent

    1.374 M + 1.911 E = 0.386 (2)

4) System of equations

       M   +         E = 0.220 (1) 1.374 M + 1.911 E = 0.386 (2)

5) Solution

From equation (1): E = 0.220 - MSubstituting in equation (2): 1.374 M + 1.911 (0.220 - M) = 0.3861.374 M + 0.4204 - 1.911 M = 0.386- 0.537 M = - 0.0344M = 0.0344 / 0.537 = 0.0641 g

 the mass of methyl alcohol (CH3OH) in the sample is 0.0641 g

H tricky one let me think and get back to u

The answer is c.  na2haso4  and na2so4