Question

given the following reaction, how much AgCl (in grams) can be formed from 25.0mL of 0.115 M AgNO3 and 45.0mL of 0.0533M MgCl2? 2AgNO39(aq) + MgCl2 (aq)--> 2AgCl(s) +Mg(NO3)2 (aq)

Answer 1

Number of moles of AgNO3 = molarity * volume of solution in L

Number of moles of AgNO3 = 0.115 * 0.025 L = 0.00288 mole

Number of moles of MgCl2 = molarity * volume of solution in L

Number of moles of MgCl2 = 0.0533 * 0.045 L = 0.00240 mol

From the balanced equation we can say that

2 mole of AgNO3 requires 1 mole of MgCl2 so

0.00288 mole of AgNO3 will require

= 0.00288 mole of AgNO3 *( 1 mole of MgCl2 / 2 mole of AgNO3)

= 0.00144 mole of MgCl2

But we have 0.00240 mole of MgCl2 which is in excess so MgCl2 is an excess reactant and AgNo3 is limiting reactant

From the balanced equation we can say that

2 mole of AgNO3 produces 2 mole of AgCl so

0.00288 mole of AgNO3 will produce

= 0.00288 mole of AgNO3 *( 2 mole of AgCl / 2 mole of AgNO3)

= 0.00288 mole of AgCl

mass of 1 mole of AgCl = 143.32 g

so the mass of 0.00288 mole of AgCl = 0.413 g

Therefore, the mass of AgCl produced would be 0.413 g