How many grams are there in 7.3 moles of NH3?

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Ответ:

littlebirdd987

16.04.2022

Chemistry

The mass of the precipitate that AgCl is 3.5803 g.

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of solute (NaCl) = 1.46 g

Molar mass of sulfuric acid = 58.5 g/mol

Volume of solution = 250 cm^3 =250 mL

1 cm^3= 1 ml

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{1.46g\times 1000}{58.5g/mol\times 250}\\\\\text{Molarity of solution}=0.09982 M$

0.09982 M is the concentration of the sodium chloride solution.

b) $NaCl (aq)+AgNO_3 (aq)\rightarrow AgCI(s)+NaNO_3(aq)$

Moles of NaCl = $\frac{1.46 g}{58.5 g/mol}=0.02495 mol$

according to reaction 1 mol of NaCl gives 1 mol of AgCl.

Then 0.02495 moles of NaCl will give:

$\frac{1}{1}\times 0.02495 mol=0.02495 mol$ of AgCl

Mass of 0.02495 moles of AgCl:

$0.02495 mol\times 143.5 g/mol=3.5803 g$

The mass of the precipitate that AgCl is 3.5803 g.

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