Question

Помогите с задачей заранее спасибо Смесь порошков магния и меди массой 12 Г обрабатывали избытком ортофосфорной кислоты. Объем выделившегося газа составил при Н.у 5,6 л. Вычислите массовую долю меди в исходной смеси.

Answer 1

Реагирует только магний составляем уавнение
3Mg+2H3PO4= Mg3(PO4)2+3H2
M(Mg)=24
m(Mg)=24*3=72
масса магния =72*5,6/22,4*3=6 г
масса меди =12-6=6 г
W(Cu)=6/12*100%=50%

Answer 2

Task: Calculating the Mass Fraction of Copper in the Original Mixture

To calculate the mass fraction of copper in the original mixture, we need to use the information provided in the question. Let's break down the problem step by step:

1. We have a mixture of magnesium and copper powders with a total mass of 12 g. 2. The mixture is treated with an excess of orthophosphoric acid, and a gas is produced. 3. The volume of the gas produced at standard temperature and pressure (N.T.P.) is 5.6 L.

To solve this problem, we need to use the concept of stoichiometry and the ideal gas law.

Step 1: Calculate the Number of Moles of Gas Produced

To calculate the number of moles of gas produced, we can use the ideal gas law equation:

PV = nRT

Where: - P is the pressure (at N.T.P., the pressure is approximately 1 atm) - V is the volume of the gas (5.6 L) - n is the number of moles of gas (what we want to find) - R is the ideal gas constant (0.0821 L·atm/mol·K)