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dedylja [7]
4 years ago
9

The two balanced equations (1) and (2) are for reactions in which gaseous carbon dioxide is produced from the combustion of (1)

solid carbon and (2) gaseous carbon monoxide.
1. C(s) + O2(g) CO2(g) + 94.0 kcal
2. CO(g) + ½ O2(g) CO2(g) + 67.6 kcal
When 112 grams of carbon monoxide are consumed according to equation 2, which of the following occurs? (atomic weights: C = 12.0 g/mol, O = 16.0 g/mol)
1.0 mole of carbon dioxide is produced.
67.6 kcal of heat are generated.
2.0 moles of oxygen are consumed.
0.25 mole of carbon dioxide is produced.
0.50 mole of oxygen is consumed.
Chemistry
1 answer:
V125BC [204]4 years ago
3 0

Answer:

2.0 mol of oxygen are consumed.

Step-by-step explanation:

You know that you will need a balanced equation with masses, moles, and molar masses, so gather all the information in one place.

M_r:   28.0                 44.0

          CO  + ½O₂ ⟶ CO₂ + 67.6 kcal

m/g:   112

<em>Step 1</em>. Convert grams of CO to moles of CO

1 mol CO = 28.0 g CO

Moles of CO = 112 × 1/28.0

<em>Step 2.</em> Convert moles of CO to moles of CO₂.

The molar ratio is 1 mol CO₂ to 1 mol CO

Moles of CO₂ = 4.000 × 1/1

Moles of CO₂ = 4.00 mol CO₂

Option A is <em>wrong</em>.

<em>Step 3.</em> Calculate the amount of heat generated.

q = ΔH

The conversion factor is 67.6 kcal/1 mol CO₂

q = 4.00 × 67.6

q = 270 kJ

Option B is <em>wrong</em>, because it gives the heat generated by 1 mol of CO.

<em>Step 4. </em>Calculate the moles of O₂ consumed

Moles of O₂ = 2.00 mol O₂

Option C is correct.

<em>Step 5.</em> Calculate the moles of CO₂ formed

Already done in Step 2.

Moles of CO₂ = 4.00 mol CO₂

Option D is <em>wrong. </em>

<em> </em>

<em>Step 6</em>. Calculate the moles of O₂ produced

Already done in Step 4.

Moles of O₂ = 2.00 mol O₂

Option E is <em>wrong. </em>

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The question is incomplete, here is the complete question:

The common laboratory solvent chloroform is often used to purify substances dissolved in it. The vapor pressure of chloroform, CHCl₃, is 173.11 mm Hg at 25°C.

In a laboratory experiment, students synthesized a new compound and found that when 7.229 grams of the compound were dissolved in 207.8 grams of chloroform, the vapor pressure of the solution was 170.51 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight of this compound?

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<u>Explanation:</u>

The equation used to calculate relative lowering of vapor pressure follows:

\frac{p^o-p_s}{p^o}=i\times \chi_{solute}

where,

\frac{p^o-p_s}{p^o} = relative lowering in vapor pressure

i = Van't Hoff factor = 1 (for non electrolytes)

\chi_{solute} = mole fraction of solute = ?

p^o = vapor pressure of pure chloroform = 173.11 mmHg

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Putting values in above equation, we get:

\frac{173.11-170.51}{173.11}=1\times \chi_A\\\\\chi_A=0.0150

This means that 0.0150 moles of compound is present in the solution

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of compound = 0.0150 moles

Given mass of compound = 7.229 g

Putting values in above equation, we get:

0.0150mol=\frac{7.229g}{\text{Molar mass of compound}}\\\\\text{Molar mass of compound}=\frac{7.229g}{0.0150mol}=481.9g/mol

Hence, the molecular mass of the compound is 481.9 g/mol

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