Answer:
Mg(s) +<em> 2</em> HCl (aq) → H₂(g) + MgCl₂
0.415g of H₂(g) <em>-Assuming mass of Mg(s) = 10.0g-</em>
Explanation:
Balancing the reaction:
Mg(s) + HCl (aq) → H₂(g) + MgCl₂
There are in products two atoms of H and Cl, the balancing equation is:
Mg(s) +<em> 2</em> HCl (aq) → H₂(g) + MgCl₂
<em>Assuming you add 10g of Mg(s) -Limiting reactant-</em>
<em />
10g of Mg are (Atomic mass: 24.305g/mol):
10g × (1 mol / 24.305g) = <em>0.411 moles of Mg</em>
<em>-Theoretical yield is the amount of product you would have after a chemical reaction occurs completely-</em>
Assuming theoretical yield, as 1 mole of Mg(s) produce 1 mole of H₂(g), theoretical yield of H₂(g) is 0.411moles H₂(g). In grams:
0.411mol H₂(g) × (1.01g / mol) = <em>0.415g of H₂(g)</em>
To get the empirical formula, we need to know the ratio of the moles of both lead and sulfur.
First, lets get the mass of each:
mass of lead = mass of evaporating dish and lead - mass of empty dish
= <span>26.927 - 25.000 = 1.927 gm of lead (Pb)
mass of sulfur = mass of dish and lead sulfide - mass of dish and lead
= </span><span>27.485 - 26.927 = 0.558 gm of sulfur (S)
The next step is to get the number of moles in 1.927 gm of Pb and in 0.558 gm of S:
From the periodic table:
molar mass of lead = </span>207.2 gm<span>
molar mass of sulfur = 32 gm
number of moles = mass / molar mass
number of moles of Pb = 1.927 / 207.2 = </span>0.0093 moles<span>
number of moles of S = 0.558 / 32 = </span><span>0.0174 moles
The third step is to get the ratio between the moles by dividing the number of moles of each by the smaller of the two:
1 mole of Pb (</span>0.0093 / 0.0093) reacts with 0.0174 / 0.0093 = 1.99 (approximately 2 moles) of sulfur.
Final step is to write the empirical formula based on the ratio:
Lead sulfide compound is written as : PbS2
The answer is ice. Nice brain teaser!
