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3 years ago

2. A solvent used in cleaning is found to contain only the elements carbon, hydrogen, and chlorine. When a

Chemistry

1 answer:

KIM [24]3 years ago

4 0

mass of C = 0.238 g

mass of H =0.00989 g

mass of Cl = 1.05 g

Explanation:

Determine the masses of carbon, hydrogen, and chlorine in the original sample of the solvent.

First we need to calculate the molar mass of carbon dioxide (CO₂) and water (H₂O)

molar mass of CO₂ = molar mass of C × 1 + molar mass of O × 2

molar mass of CO₂ = 12 × 1 + 16 × 2 = 44 g/mole

molar mass of H₂O = molar mass of H × 2 + molar mass of O × 1

molar mass of H₂O = 1 × 2 + 16 × 1 = 18 g/mole

Now, to find the mass of carbon and hydrogen in the original sample of solvent, we devise the following reasoning:

if 44 g of CO₂ contains 12 g of C

then 0.872 g of CO₂ contains X g of C

X = (0.872 × 12) / 44 = 0.238 g of C

if 18 g of H₂O contains 2 g of H

then 0.089 g of H₂O contains Y g of H

Y = (0.089 × 2) / 18 = 0.00989 g of H

And now, we can find the mass of chlorine:

mass of sample = mass of C + mass of H + mass of Cl

mass of Cl = mass of sample - mass of C - mass of H

mass of Cl = 1.30 - 0.238 - 0.00989

mass of Cl = 1.05 g

Learn more about:

combustion reaction

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Solid sodium azide (NaN3) produces solid sodium and nitrogen gas. How many grams of sodium azide are needed to yield a volume of

ch4aika [34]

Answer:

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P*V = n*R*T

In this case, the balanced reaction is:

2 NaN₃ → 2 Na + 3 N₂

You know the following about N₂:

P= 1.10 atm
V= 26.5 L
n=?

$R=0.082057 \frac{atm*L}{mol*K}$

T= 295 K

Replacing in the equation for ideal gas:

1.10 atm* 26.5 L= n* $0.082057 \frac{atm*L}{mol*K}$ *295 K

Solving:

$n=\frac{1.10 atm*26.5 L}{0.082057 \frac{atm*L}{mol*K} *295K}$

n= 1.2 moles

Now, the following rule of three can be applied: if 3 moles of N₂ are produced by stoichiometry of the reaction from 2 moles of NaN₃, 1.2 moles of N₂ are produced from how many moles of NaN₃?

$moles of NaN_{3}=\frac{1.2 molesofN_{2} *2 molesofNaN_{3} }{3 molesofN_{2} }$

moles of NaN₃= 0.8

Since the molar mass of sodium azide is 65.01 g / mol, then one last rule of three applies: if 1 mol has 65.01 grams of NaN₃, 0.8 mol how much mass does it have?

$mass of NaN_{3} =\frac{0.8 mol*65.01 grams}{1 mol}$