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ankoles [38]
3 years ago
9

5. 1.00 mol HNO3 is treated with 4.47 g of magnesium. Calculate the number of moles of

Chemistry
1 answer:
ruslelena [56]3 years ago
3 0

Answer:

The balanced equation is:

2 HNO3 + Mg ---> Mg(NO3)2 + H2

From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg

Moles of Mg:

Molar mass of Mg = 24 g/mol

Moles = Given mass / Molar Mass

Moles of Mg = 4.47 / 24 = 0.18 moles (approx)

Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed

Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3

Unreacted moles of HNO3 = Total Moles - Moles consumed

Unreacted moles of HNO3 = 0.64 moles (approx)

Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate

From the given options, we can see that 0.632 moles is the closest value to our answer

Therefore, 0.632 moles will remain after the reaction

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8 0
2 years ago
the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29
Vera_Pavlovna [14]

Answer:

ΔG = 16.218 KJ/mol

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  • dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate
  • ΔG = ΔG° - RT Ln Q

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴  R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

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