......
Cu + 4 HNO3 -->Cu(NO3)2 + 2 H2O + 2 NO2
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90.35 g Cu X (1 mol Cu/63.55 g Cu) X (2 mol H2O / 1 mol Cu) X (18.01 g H2O / 1 mol H2O) = 51.21 g H2O
keywords: the,mass,with,produced,be,water,reaction,What,above,copper,may,of,excess,shown,acid,nitric,90.35,from,What mass of water may be produced from the reaction of 90.35 g of copper with excess nitric acid shown above
