he a 1.00 m sol'n of nitric acid. how many ml of this sol'n will he need to add to the soil to completely react with 56.8 g of al(oh)3
he a?
1. meter?
add to the soil?
Please proofread and repost.
I have a question similar to this:
One of your classmates who grows hydrangeas in his garden is interested in treating his soil to grow a particular color of hydrangea. He purchases a 1.00M solution of nitric acid. How many ml of this solution will he need to add to the soil to completely react with 56.8g of aluminum hydroxide?
To determine the amount of 1.00 M nitric acid solution needed to react with 56.8 g of Al(OH)3, we can use stoichiometry and molar ratios.
The balanced chemical equation for the reaction between nitric acid (HNO3) and aluminum hydroxide (Al(OH)3) is:
2 HNO3 + Al(OH)3 -> Al(NO3)3 + 3 H2O
From the equation, we can see that 2 moles of HNO3 are required to react with 1 mole of Al(OH)3.
First, we need to calculate the number of moles of Al(OH)3 in 56.8 g. We can do this by using the molar mass of Al(OH)3.
Molar mass of Al(OH)3:
Aluminum (Al) = 26.98 g/mol
Oxygen (O) = 16.00 g/mol
Hydrogen (H) = 1.01 g/mol
Al(OH)3 = (26.98 g/mol) + 3 * ((1.01 g/mol) + (16.00 g/mol)) = 78.00 g/mol
Number of moles of Al(OH)3 = mass / molar mass = 56.8 g / 78.00 g/mol ≈ 0.727 moles
According to the balanced equation, 2 moles of HNO3 react with 1 mole of Al(OH)3.
Therefore, we would need half the number of moles of HNO3, which is 0.727 moles / 2 ≈ 0.364 moles.
Now we can calculate the volume of 1.00 M HNO3 solution needed.
Molarity (M) = moles of solute / volume of solution (in liters)
Rearranging the equation, volume of solution = moles of solute / Molarity
Volume of HNO3 solution = 0.364 moles / 1.00 mol/L = 0.364 L
Since the volume is requested in milliliters (mL), we need to convert liters to milliliters.
1 L is equivalent to 1000 mL, so 0.364 L * 1000 mL/L ≈ 364 mL
Therefore, approximately 364 mL of the 1.00 M nitric acid solution will be needed to completely react with 56.8 g of Al(OH)3.