It is easy to find the equivalent mass for this compound
.
divide the molar mass of the compound by the valence of the element/ion.
Since, Sodium shows +1 oxidation state always
and it's 2 times. so the total charge is +2
so just divide the molar mass by 2
![\text{Mass of } Na_2CO_3 = \frac{2\times \text{ mass of sodium } + \text{ Mass of Carbon } + 3\times \text{ mass of Oxygen} }{\text{total charge of Cation or Anion}} \\ \frac{2\times 23 + 12+3\times16}{2} \\ \frac{46+12+48}{2} \\ \frac{106}{2} \\ 53](https://tex.z-dn.net/?f=%20%5Ctext%7BMass%20of%20%7D%20Na_2CO_3%20%3D%20%5Cfrac%7B2%5Ctimes%20%5Ctext%7B%20mass%20of%20sodium%20%7D%20%2B%20%5Ctext%7B%20Mass%20of%20Carbon%20%7D%20%2B%203%5Ctimes%20%5Ctext%7B%20mass%20of%20Oxygen%7D%20%7D%7B%5Ctext%7Btotal%20charge%20of%20Cation%20or%20Anion%7D%7D%20%5C%5C%20%5Cfrac%7B2%5Ctimes%2023%20%2B%2012%2B3%5Ctimes16%7D%7B2%7D%20%5C%5C%20%5Cfrac%7B46%2B12%2B48%7D%7B2%7D%20%5C%5C%20%5Cfrac%7B106%7D%7B2%7D%20%5C%5C%2053%20)
Explanation:
0 10 grams that that should be the answer
N2 + 3H2 -----> 2NH3
3 mol 2 mol
When 3 mol H2 react completely 2 mol NH3 are formed.
Answer:
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