Question

Определить массу остатка которая образуется при взаимодействие 15 г BaCl2 и 10г Na2SO4

Answer 1

BaCl2+Na2SO4=BaSO4+2NaCl

n(BaCl2)=m/M=15/(137+35,5*2)=0,072 моль
n(Na2SO4)=m/M=10/(23*2+32+16*4)=0,070 моль

считаем всегда считаем по недостатку
n(Na2So4)=n(BaSO4)=0,070 моль
m=n*M=0,070*(137+32+4*16)=16,42 г

Answer 2

Calculating the Mass of the Residue

To determine the mass of the residue formed when 15g of BaCl2 reacts with 10g of Na2SO4, we can use the stoichiometry of the reaction and the molar masses of the compounds involved.

The balanced chemical equation for the reaction is: BaCl2 + Na2SO4 → BaSO4 + 2NaCl

First, let's calculate the moles of BaCl2 and Na2SO4 involved in the reaction using their respective masses and molar masses.

The molar mass of BaCl2 is 208 g/mol, and the molar mass of Na2SO4 is 142 g/mol.

Using the given masses: - Moles of BaCl2 = 15g / 208 g/mol - Moles of Na2SO4 = 10g / 142 g/mol

Let's calculate the moles of BaSO4 formed and then determine the mass of the residue.

Moles of BaSO4 formed: - From the balanced chemical equation, 1 mole of BaCl2 produces 1 mole of BaSO4. - Moles of BaSO4 = Moles of BaCl2

Now, we can calculate the mass of BaSO4 formed using the moles of BaSO4 and the molar mass of BaSO4 (233 g/mol).

Mass of BaSO4 formed: - Mass of BaSO4 = Moles of BaSO4 * Molar mass of BaSO4

Let's perform the calculations: - Moles of BaCl2 = 15g / 208 g/mol = 0.072115 moles [[2 #]] - Moles of Na2SO4 = 10g / 142 g/mol = 0.070423 moles [[2 #]] - Moles of BaSO4 formed = 0.072115 moles - Mass of BaSO4 formed = 0.072115 moles * 233 g/mol = 16.81g [[2 #]]

The mass of the residue formed when 15g of BaCl2 reacts with 10g of Na2SO4 is approximately 16.81 grams.