Calculate the mass in mg of Na+ in 10.0 g of sodium carbonate?

Question

The answer is 4.34 x 10^3, i don't know how to get to this answer, i have tried so many things maybe i am just over complicating it but please help.

Answer 1

Calculate the mass in mg of Na+ in 10.0 g of sodium carbonate?
The answer is 4.34 x 10^3

First find out the mass of Na2CO3
Mass of Na2CO3 = 2 x 23 + 12 x 1 + 3 x 16 =106 grams/mole
106 grams of Na2CO3 contains 2 Na+ ions = 2 x 23 = 46 grams of Na+

1 gram of Na2CO3 contains = 46/106 grams of Na+
10 grams of Na2CO3 contains = (46/106) x 10 grams of Na+
= 4.33962264 or 4.34 grams of Na+
= 4.34 x 10^3 milligrams of Na+

1 gram = 1000 milligram or 10^3 milligram

Answer 2

Atomic weightsof the climate in contact are Na 23 g C 12 g O sixteen o The molecular weight of Na2CO3 is 23*2+12+sixteen*3 =>106 g 106 g Na2CO3 will comprise 40 six g of Na by way of rule of three 10 g will comprise 10*40 5/106 g =>450*one thousand/106 mg

Answer 3

First find the formula for sodium carbonate: Na2CO3; then determine how many moles 10g of this is. The molecular mass of Na2CO3 is 106 g/mole, so 10g has 10/106 moles. Than means there are 10/106 moles of (Na2) and 20/106 moles of Na+. Now find the atomic mass of Na (it's 23 g/mole). Multiply the number of moles of Na times its molecular mass to get its total mass.

Answer 4

Na2CO3, right?
so there are 2 sodium ions for every molecule of sodium carbonate......
start with number of moles of Na2CO3 equivalent to 10 g
MWt is 2*(22.98977) + 12.011 + 3*(15.9994) =
105.989 g/gmole
so there are 10.0 g / 105.989 g/gmol = 0.09435 gmol
so there are
2 * 0.09435 gmole of Na+ ions = 0.1887 gmol Na+
0.1887 gmol * 23.98977 g/gmol = 4.338 g = 4338 mg
remember significant figure....10.0 that's 3 digits
4,340 mg = 4.34 * 10³ mg

Answer 5

How about we do our own homework huh? Or how else will one learn? try hard kid. good luck