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4 years ago

HELP!!!

Chemistry

1 answer:

densk [106]4 years ago

5 0

Answer:

141.0 g NaN₃.

Explanation:

It is a stichiometric problem.

Firstly, we should wright the reaction as a balanced equation:

2NaN₃ → 2Na + 3N₂

It is clear that 2.0 moles of NaN₃ decompose to 2.0 mole of Na and 3.0 moles of N₂.

Then we should convert the volume of N₂ (70.0 L) to moles via using the gas law of ideal gas: PV = nRT, n = PV / RT,

Where, P is the pressure of the gas in atm (P = 1.2 atm).

V is the volume of the gas in L(V = 70.0 L).

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature in K(T = 315 K).

∴ n of N₂ = PV / RT = (1.2 atm) (70.0 L) / (0.082 L.atm/mol.K) (315 K) = 3.252 mole.

From the stichiometry:

2.0 moles of NaN₃ decomposes to → 3.0 moles of N₂

??? mole of NaN₃ decomposes to → 3.2520 moles of N₂

The number of moles of NaN₃ = (2.0 moles of NaN₃) (3.2520 moles of N₂) / (3.0 moles of N₂) = 2.168 mole.

Finally, we can convert the number of moles of NaN₃ to mass using the relation: m = n x molar mass.

Molar mass of NaN₃ = 65.0 g/mol.

the mass of NaN₃ = n x molar mass = (2.168 mole) (65.0 g/mol) = 140.92 g ≅ 141.0 g.

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$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

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