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If 37.6 g of AgNO3 react with 28.6 g of H2SO4 according to the unbalanced equation below, what is the mass in grams of AgSO4 that could be formed?

\[ \text{AgNO}_3 \text{(aq)} + \text{H}_2\text{SO}_4 \text{(aq)} \rightarrow \text{Ag}_2\text{SO}_4 \text{(s)} + \text{HNO}_3 \text{(aq)} \]

Answer :

Final answer:

The mass of AgSO4 that could be formed is 25.59 grams.

Explanation:

To find the mass of AgSO4 that could be formed, we must first determine the limiting reactant. This is done by comparing the amount of each reactant to the stoichiometry of the balanced equation. The molar mass of AgNO3 is 169.87 g/mol and the molar mass of H2SO4 is 98.09 g/mol.

Using the molar masses, we can convert the masses of AgNO3 and H2SO4 to moles:

moles of AgNO3

= 37.6g / 169.87 g/mol

= 0.221 mol

moles of H2SO4

= 28.6g / 98.09 g/mol

= 0.292 mol

According to the stoichiometry of the balanced equation, 2 moles of AgNO3 react with 1 mole of H2SO4 to produce 1 mole of AgSO4. Therefore, the moles of AgNO3 and H2SO4 are in a 2:1 ratio.

Since we have an excess of H2SO4 (0.292 mol) compared to AgNO3, AgNO3 is the limiting reactant. Now we can calculate the moles of AgSO4 formed:

moles of AgSO4 = 0.221 mol AgNO3 x (1 mol AgSO4 / 2 mol AgNO3)

= 0.110 mol AgSO4

Finally, we can convert the moles of AgSO4 to grams:

mass of AgSO4 = 0.110 mol AgSO4 x 232.62 g/mol

= 25.59 g

Therefore, the mass of AgSO4 that could be formed is 25.59 grams.

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