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) 38.0 g H₂O B) 21.7 g H₂O C) 43.4 g H₂O D) 10.9 g H₂O E) 26.5 g H₂O

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To find the grams of water required, first calculate the moles of HNO3 formed: 75.9 g / 63.02 g/mol = 1.204 moles. From the balanced equation, 1 mole of H2O produces 2 moles of HNO3, so 1.204 moles of HNO3 requires 0.602 moles of H2O. Convert moles of H2O to grams: 0.602 moles * 18.02 g/mol = 10.9 g.

Consider the Following Balanced Reaction