Asked by Dee
How many grams of solid product can be produced by mixing 45.6 grams of chlorine gas with excess sodium metal?
Answers
Answered by
jai
first, write the balanced equation:
Cl2 (g) + 2Na (s) --> 2NaCl (s)
*chlorine gas is diatomic*
then, do dimensional analysis to cancel units:
g of NaCl = (45.6 g Cl2)*(1 mol Cl2/35.45*2 g Cl2)*(2 mol NaCl/1 mol Cl2)*(58.45 g NaCl/1 mol NaCl)
*35.45 g is atomic mass of Cl, but since it is diatomic, it should be 35.45*2 g Cl2
*23 g is atomic weight of Na, thus 23 + 35.45 = 58.45 g is weight of NaCl
solving this, we obtain 75.2 g NaCl
(always check the number of significant digits)
so there,, :)
Cl2 (g) + 2Na (s) --> 2NaCl (s)
*chlorine gas is diatomic*
then, do dimensional analysis to cancel units:
g of NaCl = (45.6 g Cl2)*(1 mol Cl2/35.45*2 g Cl2)*(2 mol NaCl/1 mol Cl2)*(58.45 g NaCl/1 mol NaCl)
*35.45 g is atomic mass of Cl, but since it is diatomic, it should be 35.45*2 g Cl2
*23 g is atomic weight of Na, thus 23 + 35.45 = 58.45 g is weight of NaCl
solving this, we obtain 75.2 g NaCl
(always check the number of significant digits)
so there,, :)
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