# How many moles of AgCl would precipitate?

Contents

As AgCl is largely insoluble in water, 0.1467 moles of AgCl will precipitate.

## How many moles of AgCl get precipitate?

Only two chloride ions are outside the coordination sphere, therefore, only two moles of AgCl would be precipitated.

## How many moles of AgCl is produced?

In order to get the mass of AgCl, you have to multiply the produced moles of AgCl, which is 0.012 moles, with the molar mass of AgCl, which is 143.32 g/mol. In doing so, the answer will be 1.72 g AgCl.

## How many moles of AgCl is precipitated when excess of AgNO3 is added to the one mole of CoCl3 4NH3 green?

How many moles of AgCl is precipitated, when excess of AgNO3 is added to the one mole of CoCl3. 4NH3 (green) : 11.

## How do you find the moles of AgCl?

Divide the given mass m by the molecular weight W of AgCl, which is given as 143 g/mole. That gives the number of moles of AgCl. Since there is one atom of Ag in each molecule of AgCl, the number of moles of Ag equals the number of moles of AgCl.

## How many moles of AgCl is precipitated when excess of AgNO3 is added?

So it gives 1 mole AgCl ppt.

## Which gives only 25 percent mole of AgCl when reacts with AgNO3?

3 NH 3 or [ Pt ( NH 3 ) 3 Cl 3 ] Cl gives only 25 % of AgCl as we can see than here 4 Cl are present , but only 1 mol of AgCl is obtained i . e . 1 4 × 100 = 25 % . This is because here the 3 Cl atoms are inside the coordination sphere while 1 Cl is outside , therefore 1 Cl is only available for AgNO 3 .

## How many moles of AgCl will be obtained when agno3?

2 moles because 2 Cl- ions are present as counter ion in the complex.

## How many grams of AgCl were produced?

0.03 moles produces (3/3) × 0.03 = 0.03 moles AgCl. Molar mass of AgCl = (108 + 35.5) g/mol = 143.5 g/mol. So, the mass of AgCl produced = (0.03 × 143.5) g = 4.305 g.

## When an excess of AgNO3 is added to the complex 3 moles of AgCl is precipitated the formula of the complex is?

When an excess of AgNO3 is added to the complex one mole of AgCl is precipitated. The formula of the complex is [CoCl2(NH3)4]Cl.

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## What is the number of moles of AgCl precipitated when one mole of cocl3 4nh3 is treated with excess of silver nitrate?

5nh3 was treated with excess of silver nitrate solution, 2 mol of agcl was precipitated.