Nitric acid,HNO3, is available ata a concentration of 16 M. how much would react with 5.58 g of KOH acording to the following equation?
HNO3+KOH-H2O+KNO3
How many moles are in 5.56 g KOH. moles = grams/molar mass
Then moles HNO3 = M x L.
To determine the amount of nitric acid (HNO3) that would react with 5.58 g of potassium hydroxide (KOH), we need to use the balanced equation and the concept of stoichiometry.
1. Start by writing down the balanced equation:
HNO3 + KOH -> H2O + KNO3
2. Calculate the molar mass of KOH:
K (39.10 g/mol) + O (16.00 g/mol) + H (1.01 g/mol) = 56.11 g/mol
3. Convert the given mass of KOH to moles. Use the molar mass of KOH:
moles of KOH = mass / molar mass
moles of KOH = 5.58 g / 56.11 g/mol ≈ 0.0993 mol
4. From the balanced equation, we can see that the stoichiometric ratio between HNO3 and KOH is 1:1.
This means that 1 mole of HNO3 reacts with 1 mole of KOH.
5. Therefore, the moles of HNO3 that would react with 0.0993 mol of KOH is also 0.0993 mol.
6. Finally, convert the moles of HNO3 to grams using the molar mass of HNO3:
grams of HNO3 = moles of HNO3 × molar mass of HNO3
grams of HNO3 = 0.0993 mol × (1 mol HNO3/1 mol) × (63.01 g/mol)
grams of HNO3 = 6.26 g
Therefore, approximately 6.26 grams of nitric acid (HNO3) would react with 5.58 g of potassium hydroxide (KOH) according to the given balanced equation.