Answer: 15.1 grams
Given reaction:
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
Mass of Na2CO3 = 20.0 g
Molar mass of Na2CO3 = 105.985 g/mol
# moles of Na2CO3 = 20/105.985 = 0.1887 moles
Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH
# moles of NaOH produced = 0.1887*2 = 0.3774 moles
Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol
Mass of NaOH produced = 0.3774*39.996 = 15.09 grams
Explanation:
The correct answer is 3) 2CO2(g) ⇄ 2CO(g) + O2(g)
this is the correct one because it is a decomposition reaction and all the number of atoms is equal on both sides.
there are 2 C atoms on both sides.
and 4 O atoms on both sides.
and 1) the atoms numbers are equal on both sides but not correct as it not a
correct number as it has 1/2 O2.
and 2) CO2(g) ⇆ CO(g) + O2
the number of O atoms is not equal on both sides of the equation.
we have 2 O atoms on the left side and 3 O atoms on the right side.
so, this not a balanced equation.
4) also not correct 2CO(g) + O2 ⇆ 2CO2
as it is not a decomposition reaction and the 2CO & O2 are as reactants not products.
so the correct answer is 3) 2CO2(g) ⇆ 2CO(g) + O2(g)
Sunlight? air? i’m not sure they could all be matter..
