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3 years ago

How many grams of NH3 are produced when 39.8 liters of N2 reacts?

Chemistry

1 answer:

algol133 years ago

3 0

Answer:

inj

Explanation:

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A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting volume was 113 mL. Calc

rodikova [14]

Explanation:

Mass of solute = 10.0 g

mass of solvent(water) = m

Volume of solvent( water) = v = 100.0 mL

Density of water= d = $1 g/cm^3=1 g/mL$

$1 mL= 1 cm^3$

$m=d\times v=1.0 g/mL\times 100.0 = 100.0 g$

Mass of solution(M) = Mass of solute + mass of solvent

M = 10.0 g + 100.0 g = 110.0 g

Volume of the solution = V = 113 mL

Density of the solution = D

$D=\frac{M}{V}=\frac{110.0 g}{113 mL}=0.9734 g/mL$

The density of the solution is 0.9734 g/ml.

Moles of phosphoric acid = $n_1=\frac{10.0 g}{98 g/mol}=0.1020 mol$

Moles of water = $n_2=\frac{100.0 g}{18g/mol}=5.556 mol$

Mole fraction of phosphoric acid = $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1020 mol}{0.1020 mol+5.556 mol}$

$\chi_1=0.01803$

Mole fraction of water = $\chi_2$

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{5.556 mol}{0.1020 mol+5.556 mol}$

$\chi_2=0.9820$

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

Moles of phosphoric acid = 0.1020 mol

Volume of the solution = V = 113 mL = 0.113 L ( 1 mL = 0.001 L)

Molarity of the solution :

$=\frac{0.1020 mol}{0.113 L}=0.903 M$

$[Molality]=\frac{\text{Moles of solute}}{\text{Mass of solvent(kg)}}$

Moles of phosphoric acid = 0.1020 mol

Mass of solvent(water) = m =100.0 g = 0.100 kg ( 1 g = 0.001 kg)

Molality of the solution :

$=\frac{0.1020 mol}{0.100 kg}=1.02 mol/kg$

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