Question

Найдите массу соли образующейся при реакции между 20 грамм оксида агния и 63 грамм азотной кислоты

Answer 1

MgO+2HNO3=Mg(NO3)2+H2O

n(MgO)=20г   /   40 г/моль=0,5моль               

n(HNO3)=63/63=1моль

n(Mg(NO3)2)=n(MgO)=0,5моль

m (Mg (NO3)2)= n*M=0,5 моль*148г/моль=74г

Ответ: Масса HNO3=74грам

Answer 2

Calculating the Mass of Salt Formed

To find the mass of salt formed in the reaction between 20 grams of silver oxide and 63 grams of nitric acid, we can use the concept of stoichiometry. The balanced chemical equation for the reaction is:

3 Ag2O + 6 HNO3 → 6 AgNO3 + 3 H2O

From the balanced equation, we can see that 3 moles of silver oxide react with 6 moles of nitric acid to produce 6 moles of silver nitrate and 3 moles of water.

Using the molar masses: - Molar mass of Ag2O = 231.735 g/mol - Molar mass of HNO3 = 63.01284 g/mol

We can calculate the moles of each reactant: - Moles of Ag2O = 20 g / 231.735 g/mol - Moles of HNO3 = 63 g / 63.01284 g/mol

Now, we can use the mole ratio from the balanced equation to find the limiting reactant and the moles of salt formed.

Let's calculate the moles of each reactant and determine the limiting reactant.

Moles of Ag2O: - Moles of Ag2O = 20 g / 231.735 g/mol = 0.0863 moles

Moles of HNO3: - Moles of HNO3 = 63 g / 63.01284 g/mol = 1 mole

Since the balanced equation shows a 3:6 mole ratio between Ag2O and HNO3, it's clear that Ag2O is the limiting reactant.

Now, we can calculate the moles of salt formed and then the mass of the salt.

Moles of salt formed: - Moles of AgNO3 = 0.0863 moles of Ag2O * (6 moles of AgNO3 / 3 moles of Ag2O) = 0.1726 moles

Mass of salt formed: - Mass of AgNO3 = 0.1726 moles * 169.873 g/mol (molar mass of AgNO3)

Mass of salt formed = 29.32 grams

Therefore, the mass of salt formed in the reaction is approximately 29.32 grams.