Final answer:
To find the number of moles of silver bromide that will form when 15.0 g of silver nitrate is reacted with excess sodium bromide, first calculate the number of moles of silver nitrate. Then, use the balanced equation to determine the mole ratio between silver nitrate and silver bromide. Finally, divide the number of moles of silver nitrate by the mole ratio to find the number of moles of silver bromide.
Step-by-step explanation:
To determine the number of moles of silver bromide that will form, we need to first calculate the number of moles of silver nitrate. From the information given, we have 15.0 grams of silver nitrate. To convert grams to moles, we divide the mass by the molar mass of silver nitrate, which is 169.88 g/mol. So, the number of moles of silver nitrate is 15.0 g / 169.88 g/mol = 0.08824 mol.
Since there is an excess of sodium bromide, all the silver nitrate will react. According to the balanced equation for the reaction:
2 AgNO3 + 2 NaBr → AgBr2 + 2 NaNO3
we can see that 2 moles of silver nitrate react to form 1 mole of silver bromide. Therefore, the number of moles of silver bromide that will form is half of the number of moles of silver nitrate: 0.08824 mol / 2 = 0.04412 mol.