Question

Please Help! Show all the actions in order. Calculate the mass of sodium nitrate (NaNO3) obtained in a reaction in which 475 mL of 12% nitric acid (HNO3)(density 1.06 g / mL) was reacted with excess sodium hydroxide(NaOH)!

Answer 1

mass of NaNO₃ : 81.515 g

Further explanation

The reaction equation is the chemical formula of reagents and product substances

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products

Reaction

HNO₃ + NaOH ⇒ NaNO₃ + H₂O

MW HNO₃ = 63 g/mol

MW NaNO₃ =85 g/mol

475 mL of 12% nitric acid (HNO₃)(density 1.06 g / mL)

Molarity HNO₃ :

`\tt M=(\%* \rho* 10)/(MW)\\\\M=(12* 1.06* 10)/(63)=2.019`

• mol HNO₃

`\tt mol=M* V\\\\mol=2.019* 0.475~L=0.959`

mol HNO₃ : mol NaNO₃ = 1 : 1 = 0.959

• mass NaNO₃

`\tt mass=mol* MW=0.959 * 85=81.515~g`