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 + AgNO₃ → 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.

Answers are 1,2,4 its correct on edgenuity.

The answer will be c