The balanced equation shows how sodium chloride reacts with silver nitrate to form sodium nitrate and silver chloride. NaCl + AgNO3 NaNO3 + AgCl If 4.00 g of NaCl react with 10.00 g of AgNO3, what is the excess reactant

The excess reactant is the NaCl.

Explanation:

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

This is a precipitation reaction.

We need to convert the mass of each reactant to moles:

4 g / 58.45 g/mol = 0.068 moles of NaCl

10 g / 169.87 g/mol = 0.059 moles of nitrate.

As stoichiometry is 1 by 1, 1 mol of chloride will react with 1 mol of nitrate.

If I have 0.059 moles of nitrate, I will need the same amount of chloride and I have 0.068 moles. I still have chloride, therefore the excess reactant is the NaCl.

0.068 mol - 0.059 mol = 0.009 moles are the moles of NaCl that remains after the reaction is complete

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

i believe the answer is three water molecules