NaCl + AgNO3 - > AgCl + NaNO3

How much AgNO3 is consumed during the reaction with 78 g of NaCl?

Answer is: 226,88 grams of AgNO₃ is consumed.

Chemical reaction: NaCl + AgNO ₃ → AgCl + NaNO₃.

m (NaCl) = 78 g.

n (NaCl) = m (NaCl) : M (NaCl).

n (NaCl) = 78 g : 58,4 g/mol.

n (NaCl) = 1,335 mol.

From chemical reaction: n (NaCl) : n (AgNO₃) = 1 : 1.

n (AgNO₃) = 1,335 mol.

m (AgNO₃) = 1,335 mol · 169,87 g/mol.

m (AgNO₃) = 226,88 g.