How many grams of precipitate can be formed in the reaction between 6.00 ml of 0.10 m silver nitrate and 5.00 ml of 0.15 m potassium chloride? hint: start with a balanced chemical equation. do not enter units with your answer.

0.09 g of precipitate can be formed.

Explanation:

The chemical equation can be written as follows:

AgNO₃ + KCl → AgCl(↓) + NO₃⁻ + K⁺

From the equation, we know that 1 mol AgNO₃ reacts with 1 mol KCl to produce 1 mol AgCl.

The problem gives us data to calculate the initial number of moles of silver nitrate and potassium chloride:

n° of moles of silver nitrate = concentration * volume

n° of moles of silver nitrate = 0.10 mol/l * 0.006 l = 6 x 10⁻⁴ mol AgNO₃

n° of moles of KCl = 0.15 mol/l * 0.005 l = 7.5 x 10⁻⁴ mol KCl

Since 1 mol AgNO₃ reacts with 1 mol KCl, 6 x 10⁻⁴ mol AgNO₃ will react with 6 x 10⁻⁴ mol KCl and produce 6 x 10⁻⁴ mol AgCl.

1.5 x 10⁻⁴ mol KCl is in excess.

The molar mass of AgCl is 143.32 g/mol, then, 6 x 10⁻⁴ mol AgCl will have a mass of (6 x 10⁻⁴ mol AgCl * 143.32 g / 1 mol) 0.09 g.

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number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.