Okay I did the math and I'm guessing around 18*C
Answer:
4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium
Explanation:
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:
- HCl: 2 moles
- Na: 1 mole
- NaCl: 2 moles
- H₂: 1 mole
You know the following masses of each element:
- H: 1 g/mole
- Cl: 35.45 g/mole
- Na: 23 g/mole
So, the molar mass of each compound participating in the reaction is:
- HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
- Na: 23 g/mole
- NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
- H₂: 2* 1 g/mole= 2 g/mole
Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:
- HCl: 2 moles* 36.45 g/mole= 72.9 g
- Na: 1 mole* 23 g/mole= 23 g
- NaCl: 2 moles* 58.45 g/mole= 116.9 g
- H₂: 1 mole* 2 g/mole= 2 g
So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

mass of H₂= 4.8 g
<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>
Answer:- 1840 g.
Solution:- We have been given with 3.35 moles of and asked to calculate it's mass.
To convert the moles to grams we multiply the moles by the molar mass of the compound. Molar mass of the compound is the sum of atomic masses of all the atoms present in it.
molar mass of = atomic mass of Hg + 2(atomic mass of I) + 6(atomic mass of O)
= 200.59+2(126.90)+6(16.00)
= 200.59+253.80+96.00
= 550.39 gram per mol
Let's multiply the given moles by the molar mass:

= 1843.8 g
Since, there are three sig figs in the given moles of compound, we need to round the calculated my to three sig figs also. So, on rounding off to three sig figs the mass becomes 1840 g.
Um i think gold... i think?