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Sodium metal and water react to form hydrogen and sodium hydroxide. if 17.94 g of sodium react with water to form 0.78 g of hydrogen and 31.20 g of sodium hydroxide, what mass of water was involved in the reaction

Respuesta :

kenmyna
The mass  of water  involved  in reaction is  calculated   as  follows

in  a  reaction mass  of reactants  should  =  mass  of  products 

mass  of  product = mass of hydrogen+  mass of    sodium   hydroxide
= 0.78 g + 31.20  g  =  31.98  grams

mass  of reactant = mass  of sodium metal+ mass of water
let the  mass of water be  represented  by  y
=17.94g  +y  =  31.98 g
like  terms together

y=31.98 g -17.94 g =14.04  grams  of water

The mass of water involved in the reaction = 14.04 g

let's write the chemical equation and balance it.

Na + H₂O → H₂ + NaOH

Balancing the equation

2Na + 2H₂O → H₂ + 2NaOH

Using the law of conservation of mass, the mass of the reactant should be equal to the mass of the product. Therefore,

let

the mass of water = x

17.94g + x g = 0.78 g + 31.20 g

x = 31.98 - 17.94

x = 14.04 g

Mass of water that react with sodium = 14.04 g

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