Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is excess NO2 present. The molar masses are as follows: H2O = 18.02 g/mol, HNO3 = 63.02 g/mol.3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

• A. 20.00 g
• B. 15.00 g
• C. 10.86 g
• D. 5.43 g

Answer: (C)

Verified step by step guidance

1. Identify the balanced chemical equation: 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g). This tells us the stoichiometric relationship between the reactants and products.
2. Determine the molar mass of the substances involved. For H2O, it is 18.02 g/mol, and for HNO3, it is 63.02 g/mol.
3. Calculate the moles of HNO3 needed using its molar mass. Use the formula: moles of HNO3 = mass of HNO3 / molar mass of HNO3. Substitute the given mass of HNO3 (75.9 g) and its molar mass (63.02 g/mol) into the formula.
4. Use the stoichiometry of the balanced equation to find the moles of H2O required. According to the equation, 2 moles of HNO3 are produced per 1 mole of H2O. Therefore, calculate the moles of H2O using the ratio: moles of H2O = (moles of HNO3) * (1 mole H2O / 2 moles HNO3).
5. Convert the moles of H2O to grams using its molar mass. Use the formula: mass of H2O = moles of H2O * molar mass of H2O. Substitute the calculated moles of H2O and the molar mass of H2O (18.02 g/mol) into the formula to find the mass of water required.