Respuesta :
When 31.2 ml of 0.500 M AgNO₃ is added to 25.0 ml of 0.300 M NH₄Cl, grams of AgCl are formed is 1.074 g.
The reaction is given as:
AgNO₃ + NH₄Cl ----> AgCl + NH₄NO₃
given that :
molarity of AgNO₃ = 0.500 M
volume = 31.2 mL = 0.0312 L
moles of AgNO₃ = 0.500 × 0.0312
= 0.0156 mol
moles of NH₄Cl = molarity × volume
= 0.300 × 0.025
= 0.0075 mol
1 mole of NH₄Cl = 1 mole of AgCl
mass of AgCl = moles × molar mass
= 0.0075 × 143.32
= 1.074 g
Thus, When 31.2 ml of 0.500 M AgNO₃ is added to 25.0 ml of 0.300 M NH₄Cl, grams of AgCl are formed is 1.074 g.
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