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3 years ago

The reaction described in part a required 3.65 l of sodium chloride. what is the concentration of this sodium chloride solution?

1 answer:

7 0

Chemical reaction: AgNO₃ + NaCl → AgCl + NaNO₃.
Missing question: Part A:What mass of silver chloride can be produced from 1.40 L of a 0.180 M solution of silver nitrate?
V(AgNO₃) = 1,4 L.
c(AgNO₃) = 0,180 M = 0,180 mol/L.
V(NaCl) = 3,65 L.
n(AgNO₃) = V(AgNO₃) · c(AgNO₃).
n(AgNO₃) = 1,4 L · 0,180 mol/L.
n(AgNO₃) = 0,252 mol.
From chemical reaction: n(AgNO₃) : n(NaCl) = 1 : 1.
n(NaCl) = 0,252 mol.
c(NaCl) = 0,252 mol ÷ 3,65 L.
c(NaCl) = 0,069 mol/L

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Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

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For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

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To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

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