Question

If 0.205 moles of AgNO3, react

with 0.155 moles of H2SO4
according to this UNBALANCED
equation below, what is the mass
in grams of Ag2SO4, that could be
formed?
AgNO (aq) + H₂SO (aq) → Ag,SO (s) + HNO, (aq)

Answer 1

The balanced equation for the reaction is:

2AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2HNO3(aq)

From the equation, we see that 2 moles of AgNO3 react with 1 mole of H2SO4 to produce 1 mole of Ag2SO4. Therefore, the limiting reagent is H2SO4, as it has the smaller number of moles.

The balanced equation shows that 1 mole of H2SO4 produces 1 mole of Ag2SO4, so 0.155 moles of H2SO4 will produce 0.155 moles of Ag2SO4.

The molar mass of Ag2SO4 is:

2(Ag) + S + 4(O) = 2(107.87 g/mol) + 32.06 g/mol + 4(16.00 g/mol) = 311.99 g/mol

Therefore, the mass of Ag2SO4 that could be formed is:

0.155 moles × 311.99 g/mol = 48.36 g

So, the mass in grams of Ag2SO4 that could be formed is 48.36 g.

Answer 2

The balanced equation is:

2AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2HNO3(aq)

From the balanced equation, 2 moles of AgNO3 react with 1 mole of H2SO4 to form 1 mole of Ag2SO4.

Given that 0.205 moles of AgNO3 react with 0.155 moles of H2SO4, we can calculate the limiting reagent as follows:

1 mole of H2SO4 reacts with (2 moles of AgNO3 / 1 mole of H2SO4) = 2 moles of AgNO3

Therefore, H2SO4 is the limiting reagent since there are only 0.155 moles of it, which is less than the required 0.205 moles of AgNO3.

Using the amount of H2SO4, we can calculate the amount of Ag2SO4 formed:

0.155 moles of H2SO4 × (1 mole of Ag2SO4 / 2 moles of AgNO3) × (2 × 107.87 g/mol) = 16.95 g

Therefore, the mass of Ag2SO4 that could be formed is 16.95 grams.

I Hope This Helps!