Question

If 21.7 mL Na, with a density of 0.97 g/mL, react with excess oxygen gas, how many moles of sodium oxide can be formed?

unbalanced equation: Na + O2 “yields”/ Na2O

0.458 mol Na2O

0.487 mol Na2O

1.83 mol Na2O

1.95 mol Na2O

@ganeshie8

Answer 1

@SolomonZelman do you happen to do any chem?

Answer 2

2Na + O2 --> Na2O

21.7 *.97 = 21.05 g
21.05/23 = .915 moles of Na
===> .915/2 = 0.458 ,oles of Na2O

Answer 3

Na+O2-->Na2O
Balance the chemical equation
Write down the number of atoms that you have on each side of the equation. Look at the subscript next to each atom to find the number of atoms in the equation.
On the reactant Side: 1 Na 2 O
On the product side: 2 Na 1 O
Balance the O by adding a coefficient of 2 to Na2O
Na+O2-->2Na2O
On the reactant Side: 1 Na 2 O
On the product side: 4 Na 2 O
Now add a coefficient of 4 to Na on the reactant side.
4Na+O2-->2Na2O
Look at the mole ratio according to the balanced chemical equation
4 moles of sodium are required to react with 1 mole of oxygen to produce 2 moles of sodium oxide
4:1:2
Density=mass(g)/volume(ml)
0.97=mass(g)/21.7(ml)
mass(g) of Na=0.97*21.7
Number of moles=Mass(g)/Molar Mass(g/mol)
Na:23 g (atomic weight in the periodic table)
21.049/23=Number of moles of Na

Answer 4

0.92 moles of Na
4:2
0.92:x moles of Na2O
x=0.46 moles of Na2O

Answer 5

Good work Abmon!

Answer 6

thank you @Preetha :)