Final answer:
D) 100.0 mL.To determine the volume of 1.50 M Nitric acid needed to react with 100.0 g of cuprous oxide, you can use stoichiometry and the balanced chemical equation.
Step-by-step explanation:
To find out how many milliliters of 1.50 M Nitric acid is required to react with 100.0 g of cuprous oxide, we need to use stoichiometry and the balanced chemical equation. The balanced chemical equation for the reaction between Nitric acid (HNO3) and cuprous oxide (Cu2O) is:
Cu2O + 4HNO3 → 2Cu(NO3)2 + 2H2O
From the balanced equation, we can see that the ratio of HNO3 to Cu2O is 4:1. Thus, for every 4 moles of HNO3 used, 1 mole of Cu2O will be reacted. To calculate the volume of 1.50 M Nitric acid needed, we can use the equation:
moles HNO3 = (volume HNO3 in L) x (concentration of HNO3 in M)
The number of moles of HNO3 needed to react with 100.0 g of Cu2O can be calculated using the molar mass of Cu2O and the molar mass of HNO3. Finally, the volume of 1.50 M Nitric acid needed can be calculated by rearranging the moles HNO3 equation.
The volume can be calculated by determining the number of moles of HNO3 needed to react with Cu2O and then converting it to milliliters based on the concentration of the Nitric acid solution.