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kramer
4 years ago
12

When calcium carbide (CaC2) comes into contact with water, it produces ethyne (C2H2) according to this equation: CaC2(s) + 2H2O(

l) → C2H2(g) + Ca(OH)2(s). Suppose this reaction takes place in a sealed container in which the gas occupies a volume of 1.50 liters. The container is kept at a temperature of 0°C. If 5.18 grams of calcium carbide reacts with excess water, what is the partial pressure of the ethyne gas produced, in atmospheres? Answer: 1.21
Chemistry
1 answer:
SOVA2 [1]4 years ago
7 0

Answer:

1.21

Explanation:

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A compound has a molecular weight of 146 g/mol. A 0.3250 g sample of the compound contains 0.1605 g of carbon, 0.0220 g of hydro
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Answer: The molecular formula is C_6H_{10}S_2

Explanation:

We are given:

Mass of C = 0.1605 g

Mass of H= 0.0220 g

mass of S = 0.1425 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{0.1605g}{12g/mole}=0.0134moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.0220g}{1g/mole}=0.0220moles

Moles of S =\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{0.1425g}{32g/mole}=0.0044moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{0.0134}{0.0044}=3

For H = \frac{0.0220}{0.0044}=5

For S =\frac{0.0044}{0.0044}=1

The ratio of C : H: S=  3: 5: 1

Hence the empirical formula is C_3H_5S

The empirical weight of C_3H_5S = 3(12)+5(1)+1(32)= 73g.

The molecular weight = 146 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{146}{73}=2

The molecular formula will be=2\times C_3H_5S=C_6H_{10}S_2

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