How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?

The complete balanced chemical reaction is:

2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S

First let us calculate the number of moles of AgNO3.

moles AgNO3 = 0.315 M * 0.035 L

moles AgNO3 = 0.011025 mol

From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:

moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)

moles Na2S required = 5.5125 x 10^-3 mol

Therefore volume required is:

volume Na2S = 5.5125 x 10^-3 mol / 0.260 M

volume Na2S = 0.0212 L = 21.2 mL

if you mean hydrogen and oxygen then water is the product