Chemistry, 18.09.2019 06:10 Bayabbay8380

If 7.81 g of mg undergoes the following reaction what is the wyield if 6.76 g of magnesium nitride is produced?
3 mg(s) + nale) - mg3n2 (5).

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Answer from: Aspect148

Corrected question:

Since the question has some errors, I copy it with the corrections:

If 7.81 g of Mg undergoes the following reaction what is the % yield, if 6.76 g of magnesium nitride is produced?

3 Mg(s) + N₂ (g) → Mg₃N₂ (s).62.6 %

Explanation:

1) Mole ratios:

The coefficients on the balanced chemical equation give the mole ratios:

3 mol Mg(s) : 2 mol N₂ (g) : Mg₃N₂

2) Theoretical yield:

Number of moles of Mg that react, n:

n = mass in grams / molar mass = 7.81 g / 24.305 g/mol = 0.321 mol

Number of moles of magnesium nitride that can be produced

Proportion: 3 mol Mg / 1 mol Mg₃N₂ = 0.321 mol Mg / x

⇒ x = 0.321 mol Mg × 1 mol Mg₃N₂ / 3 mol Mg = 0.107 mol Mg₃N₂

mass = molar mass × number of moles = 100.9494 g/mol × 0.107 mol = 10.80 g

3) Percent yield, %

% = (actual yield / theoretical yield) × 100 = (6.76 g / 10.80 g) × 100 = 62.6%

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

when the volume of gas is changed from 717.73 ml to 852 ml the temperature will change from 315 c to 452 c

calculation

this is calculated using the charles law formula

that is, v1/t1 = v2/t2

v1=?

t1 = 315 c into kelvin = 315 +273 = 588 k

v2 = 852 ml

t2 = 425c + 273 =698 k

make v1 the subject of the formula ( by multiplying both side by t1)

v1=( v2 xt1) / t2

v1= (852 ml x 588 k)/ 698 k= 717.73 ml

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If 7.81 g of mg undergoes the following reaction what is the wyield if 6.76 g of magnesium nitride i...

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