1. Use your value for the heat of neutralization,H 1 =-158.0kJ/mol, to determine

Question

1. Use your value for the heat of neutralization,H1=-158.0kJ/mol, to determine the amount of heatevolved when 20.00mL of 0.4012M HNO3 reacts with 20.00mLof 0.3978 NaOH. 1. Use your value for the heat of neutralization,H1=-158.0kJ/mol, to determine the amount of heatevolved when 20.00mL of 1.984M HNO3 reacts with 20.00mLof 2.016M NaOH. 1. Use your value for the heat of neutralization,H1=-158.0kJ/mol, to determine the amount of heatevolved when 20.00mL of 0.4012M HNO3 reacts with 20.00mLof 0.3978 NaOH. 1. Use your value for the heat of neutralization,H1=-158.0kJ/mol, to determine the amount of heatevolved when 20.00mL of 1.984M HNO3 reacts with 20.00mLof 2.016M NaOH.

Explanation / Answer

1) HNO3 + NaOH NaNO3 + H2O 20 mL *0.4012 M = 8.02400 mmol HNO3 20 mL*(0.3978 M) = 7.95600 mmol NaOH 7.956 mmol react q = 158.0 kJ/mol *(7.956 e-3 mol) = 1.257 kJ 2) 20 mL *(1.984 M) *(1e-3 mol/1mmol) *(158 kJ) = 6.269 kJ multiply by 1.984 because this is the concentration of the limitingreactant (lower concentration and 1:1 stoichiometry between acidand base)

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