Final answer:
When magnesium reacts with sulfuric acid, the products are magnesium sulfate and hydrogen. If there are 15 grams of magnesium at the start of the reaction, approximately 240.74 grams of magnesium sulfate will be present.
Step-by-step explanation:
When magnesium reacts with sulfuric acid, the products are magnesium sulfate and hydrogen. The balanced chemical equation for this reaction is:
2Mg + H2SO4 → MgSO4 + H2
From the balanced equation, we can see that for every 2 moles of magnesium that react, 1 mole of magnesium sulfate is produced. Therefore, the amount of magnesium present in the magnesium sulfate can be calculated using the ratio of their molar masses. The molar mass of magnesium is 24.31 g/mol, and the molar mass of magnesium sulfate (MgSO4) is 120.37 g/mol.
Let's calculate the mass of magnesium sulfate produced:
2 moles of magnesium sulfate = 2 * 120.37 g = 240.74 g
Now, we can use the following ratio:
Magnesium sulfate produced : Magnesium present = 240.74 g : 15 g
Cross-multiplying,
Magnesium sulfate produced = (240.74 g * 15 g) / 15 g = 240.74 g
Therefore, 240.74 grams of magnesium sulfate will be present if there are 15 grams of magnesium at the start of the reaction.