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3 years ago

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha

t mass of precipitate will form if 1.50 L of highly concentrated Pb ( ClO 3 ) 2 is mixed with 0.200 L 0.200 M NaI ? Assume the reaction goes to completion.

1 answer:

Citrus2011 [14]3 years ago

6 0

<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles$

The chemical equation for the reaction of NaI and lead chlorate follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = $\frac{1}{2}\times 0.04=0.02mol$ of lead iodide

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

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

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

$0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g$

Hence, the mass of lead iodide produced is 9.22 grams

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3 years ago

50.g of NaNO3 was dissolved in 1250 mL of water. what is the molality of the solution? [ Molar mass of NaNO3 = 85 g/mol

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Approximately $0.47\; \rm mol \cdot L^{-1}$ (note that $1\; \rm M = 1 \; \rm mol \cdot L^{-1}$ .)

Explanation:

The molarity of a solution gives the number of moles of solute in each unit volume of the solution. In this $\rm NaNO_3$ solution in water,

Let $n$ be the number of moles of the solute in the whole solution. Let $V$ represent the volume of that solution. The formula for the molarity $c$ of that solution is:

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In this question, the volume of the solution is known to be $1250\; \rm mL$ . That's $1.250\; \rm L$ in standard units. What needs to be found is $n$ , the number of moles of $\rm NaNO_3$ in that solution.

The molar mass (formula mass) of a compound gives the mass of each mole of units of this compound. For example, the molar mass of $\rm NaNO_3$ is $85\; \rm g \cdot mol^{-1}$ means that the mass of one mole of

$\displaystyle n = \frac{m}{M}$ .

For this question,

$\begin{aligned}&n\left(\mathrm{NaNO_3}\right) \\ &= \frac{m\left(\mathrm{NaNO_3}\right)}{M\left(\mathrm{NaNO_3}\right)}\\&= \frac{50\; \rm g}{85\; \rm g \cdot mol^{-1}} \\& \approx 0.588235\; \rm mol\end{aligned}$ .

Calculate the molarity of this solution:

$\begin{aligned}c &= \frac{n}{V} \\&= \frac{0.588235\; \rm mol}{1.250\; \rm L} \\&\approx 0.47\;\rm mol \cdot L^{-1}\end{aligned}$ .

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4 years ago

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Choice B. The solid with hydrogen bonding.

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<h3>Further explanation</h3>

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