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tangare [24]
4 years ago
13

Calcium nitrate tetrahydrate dissolves in cold water to the extent of 266 g per 100 cm3. What is the concentration of nitrate io

ns in this solution? a. 32.4 M b. 22.5 M c. 11.3 M d. 16.2 M e. 8.10 M
Chemistry
1 answer:
Soloha48 [4]4 years ago
4 0

Answer : The correct option is, (b) 22.5 M

Explanation : Given,

Mass of calcium nitrate tetrahydrate = 266 g

Molar mass of calcium nitrate tetrahydrate  = 236.15 g/mole

Volume of solution = 100cm^3=100mL

Molarity : It is defined as the moles of solute present in one liter of solution.

Formula used :

Molarity=\frac{\text{Mass of calcium nitrate tetrahydrate }\times 1000}{\text{Molar mass of calcium nitrate tetrahydrate }\times \text{volume of solution in ml}}

Now put all the given values in this formula, we get:

Molarity=\frac{266g\times 1000}{236.15g/mole\times 100ml}=11.26mole/L=11.26M

As calcium nitrate tetrahydrate dissociate to give 1 mole of calcium ion, 2 moles of nitrate ion and 4 moles of water.

The concentration of nitrate ion = 2\times 11.26M=22.5M

Thus, the concentration of nitrate ion is, 22.5 M

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Answer:

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<em>1) To determine the limiting reactant of the reaction:</em>

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The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.

The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.

  • From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).

The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).

∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>

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