Question

M (AgNO3)=46г m (AgCl) - ?

Answer 1

Ответ:

Объяснение:

2AgNO3+CaCl2=2AgCl+Ca(NO3)2

M(AgNO3)=108+14+48=170 г/моль

n(AgNO3)=m(AgNO3)/M(AgNO3)=17г:170 г/моль=0,1 моль

n(AgCl)=2/2n(AgNO3)=nAgNO3=0,1 моль

M(AgCl)=108+35,5=143,5 г/моль

m(AgCl)=0,1 моль*143,5 г/моль=14,35 г.

Answer 2

To calculate the mass of silver chloride (AgCl) formed when 46 grams of silver nitrate (AgNO3) reacts, we need to know the balanced chemical equation for the reaction between silver nitrate and chloride ions (Cl-). The balanced chemical equation for the reaction is:

AgNO3 + NaCl -> AgCl + NaNO3

From the equation, we can see that 1 mole of AgNO3 reacts with 1 mole of NaCl to produce 1 mole of AgCl.

Now, we need to find the molar mass of AgCl:

Ag: 1 atom * 107.87 g/mol = 107.87 g/mol Cl: 1 atom * 35.45 g/mol = 35.45 g/mol

Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol

Next, we calculate the number of moles of AgNO3:

Number of moles = Mass / Molar mass Number of moles of AgNO3 = 46 g / 169.87 g/mol ≈ 0.2706 mol

Since the stoichiometric ratio of AgNO3 to AgCl is 1:1, the number of moles of AgCl formed will also be approximately 0.2706 mol.

Finally, we find the mass of AgCl:

Mass of AgCl = Number of moles * Molar mass Mass of AgCl ≈ 0.2706 mol * 143.32 g/mol ≈ 38.76 g

So, approximately 38.76 grams of AgCl will be formed when 46 grams of AgNO3 reacts.