Step-by-step explanation:
First we have to balance this equation:
__ AgNO₃ (aq) + __ H₂SO₄ (aq) ---> __ Ag₂SO₄ (s) + __ HNO₃ (aq)
If we look at the equation we can see that we have two atoms of Ag on the right side of the equation and just one atom of Ag on the left side. So we can change the coefficient for AgNO₃ and write a 2 there.
Also, we have 2 atoms of H on the left side of the equation and just one on the right side, so we can change the coefficient for HNO₃ and write a 2 there. The balanced equation is:
2 AgNO₃ (aq) + H₂SO₄ (aq) ---> Ag₂SO₄ (s) + 2 HNO₃ (aq)
Now, 48.7 g of AgNO₃ will react with 28.6 g of H₂SO₄. We have to determine the limiting reagent. To do that we can find the number of moles of Ag₂SO₄ produced by those grams.
First we can convert those masses into moles using their molar masses.
molar mass of Ag = 107.87 g/mol
molar mass of N = 14.01 g/mol
molar mass of O = 16.00 g/mol
molar mass of H = 1.01 g/mol
molar mass of S = 32.07 g/mol
molar mass of AgNO₃ = 107.87 g/mol + 14.01 g/mol + 3 * 16.00 g/mol
molar mass of AgNO₃ = 169.88 g/mol
molar mass of H₂SO₄ = 2 * 1.01 g/mol + 32.07 g/mol + 4 * 16.00 g/mol
molar mass of H₂SO₄ = 98.09 g/mol
mass of AgNO₃ = 48.7 g
mass of H₂SO₄ = 28.6 g
moles of AgNO₃ = 48.7 g * 1 mol/(169.88 g)
moles of AgNO₃ = 0.287 moles
moles of H₂SO₄ = 28.6 g * 1 mol/(98.09 g)
moles of H₂SO₄ = 0.292 moles
2 AgNO₃ (aq) + H₂SO₄ (aq) ---> Ag₂SO₄ (s) + 2 HNO₃ (aq)
We have to determine the limiting reactant of this reaction. We mixed 0.287 moles of AgNO₃ with 0.292 moles of H₂SO₄. We can find the number of moles of Ag₂SO₄ produced by each reactant with an excess of the other one.
According to the coefficients of the reaction the molar ratio between AgNO₃ and Ag₂SO₄ is 2 to 1.
2 moles of AgNO₃ = 1 mol of Ag₂SO₄
moles of Ag₂SO₄ = 0.287 moles of AgNO₃ * 1 mol of Ag₂SO₄/(2 moles of AgNO₃)
moles of Ag₂SO₄ = 0.144 moles
According to the coefficients of the reaction the molar ratio between H₂SO₄ and Ag₂SO₄ is 1 to 1.
1 mol of H₂SO₄ = 1 mol of Ag₂SO₄
moles of Ag₂SO₄ = 0.292 moles of H₂SO₄ * 1 mol of Ag₂SO₄/(1 mol of H₂SO₄)
moles of Ag₂SO₄ = 0.292 moles
So, 0.287 moles of AgNO₃ (when reacting with enough H₂SO₄) will produce 0.144 moles of Ag₂SO₄. And 0.292 moles of H₂SO₄ (when reacting with enough AgNO₃) will produce 0.292 moles of Ag₂SO₄. So, AgNO₃ is limiting our reaction and H₂SO₄ is in excess.
We found that 0.144 moles of Ag₂SO₄ are produced and finally we can convert that into moles using the molar mass of Ag₂SO₄.
molar mass of Ag = 107.87 g/mol
molar mass of O = 16.00 g/mol
molar mass of S = 32.07 g/mol
molar mass of Ag₂SO₄ = 2 * 107.87 g/mol + 32.07 g/mol + 4 * 16.00 g/mol
molar mass of Ag₂SO₄ = 311.81 g/mol
mass of Ag₂SO₄ = 0.144 moles * 311.81 g/mol
mass of Ag₂SO₄ = 44.9 g
Answer: The mass of Ag₂SO₄ that could be formed is 44.9 g.