30.0 mL of 0.50 M NaOH to neutralize 15.0 mL of HNO3, what is the concentration (molarity) of the HNO3?
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The reaction is as follows:
2 H₂(g) + O₂(g) → 2 H₂O(g) . ΔH = - 483.5 kJ
Using the change in enthalpy and heat, calculate the moles as follows:
Moles of H₂ = -
x 2 mol H₂
= - 216 kJ / (-483.5 kJ) x 2 mol H₂
= 0.893 mol H₂
Calculate the mass of H₂ using the moles and molar mass as follows:
0.893 mol H₂ x (2.02 g H₂ / 1 mol H₂) = 1.79 g H₂
Therefore, the mass of hydrogen gas is 1.79 g
2 H₂(g) + O₂(g) → 2 H₂O(g) . ΔH = - 483.5 kJ
Using the change in enthalpy and heat, calculate the moles as follows:
Moles of H₂ = -
= - 216 kJ / (-483.5 kJ) x 2 mol H₂
= 0.893 mol H₂
Calculate the mass of H₂ using the moles and molar mass as follows:
0.893 mol H₂ x (2.02 g H₂ / 1 mol H₂) = 1.79 g H₂
Therefore, the mass of hydrogen gas is 1.79 g