Question

if 0.683 grams of silver chloride is produced how much (mass) silver nitrate would need to be reacted

Answer 1

To produce 0.683 grams of silver chloride, approximately 0.808 grams of silver nitrate would need to be reacted.

Step-by-step explanation:

The question is asking for the mass of silver nitrate that would need to be reacted in order to produce 0.683 grams of silver chloride. To determine this, we need to use the balanced chemical equation for the reaction between silver nitrate and sodium chloride:

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

From the equation, we can see that the stoichiometric ratio between silver nitrate and silver chloride is 1:1. This means that for every 1 mole of silver chloride produced, 1 mole of silver nitrate is consumed. To find the mass of silver nitrate, we can use the molar mass of silver nitrate:

Molar mass of AgNO3 = 169.88 g/mol

To calculate the mass of silver nitrate needed, we can set up the following equation:

(mass of silver nitrate)/(molar mass of AgNO3) = (mass of silver chloride)/(molar mass of AgCl)

Solving for the mass of silver nitrate:

(mass of silver nitrate) = (mass of silver chloride) × (molar mass of AgNO3)/(molar mass of AgCl)

Substituting the given values:

(mass of silver nitrate) = 0.683 g × (169.88 g/mol)/(143.32 g/mol)

(mass of silver nitrate) ≈ 0.808 g

Answer 2

The (mass) silver nitrate would needed to be reacted is calculated as follow

by assuming that AgNo3 reacted with Bacl

that is 2AgNo3 + BaCl2 ---> 2Agcl + Ba(No3)2

find the moles AgNo3 = 0.683g/169.87 g/mol =4.02 x10^-3 moles

by use of mole ratio between Agcl and Agno3 which is 2:2 this implies that the moles of AgNO3 also 4.02 x10^-3

mass= moles x molar mass

= (4.02 x10^-3) x 143.37= 0.576 grams