Reason 1: Allows us to handle larger numbers with ease
Reason 2: Allows us to easily tell the number of significant figures with ease.
The balanced net equation for
BaCl2 (aq) + H2SO4(aq) → BaSO4(s) + HCl (aq) is
Ba^2+(aq) +SO4^2- → BaSO4 (s)
<u><em>Explanation</em></u>
Ionic equation is a chemical equation in which electrolytes in aqueous solution are written as dissociated ions.
<u>ionic equation is written using the below steps</u>
Step 1: <em>write a balanced molecular equation</em>
BaCl2 (aq) +H2SO4 (aq)→ BaSO4(s) +2HCl (aq)
Step 2: <em>Break all soluble electrolytes in to ions</em>
= Ba^2+ (aq) + 2Cl^-(aq) + 2H^+(aq) + SO4^2-(aq)→ BaSO4(s) + 2H^+(aq) +2Cl^- (aq)
step 3: <em>cancel the spectator ions in both side of equation ( ions which do not take place in the reaction)</em>
<em> </em><em> =</em> 2Cl^- and 2H^+ ions
Step 4: <em>write the final net equation</em>
<em> Ba^2+(aq) + SO4^2-(aq)→ BaSO4(s</em><em>)</em>
The answer to this would be 22 mL
Answer:
1. The products of this reaction are ZnCl₂ and H₃PO₄.
2. 14.57 g.
Explanation:
<em>1. What would the products of this reaction be?</em>
- The balanced reaction between Zn₃(PO₄)₂ and HCl is represented as:
<em>Zn₃(PO₄)₂ + 6HCl → 3ZnCl₂ + 2H₃PO₄,</em>
It is clear that 1.0 mol of Zn₃(PO₄)₂ reacts with 6.0 mol of HCl to produce 3.0 mol of ZnCl₂ and 2.0 mol of H₃PO₄.
So, the products of this reaction are ZnCl₂ and H₃PO₄.
<em>2. If we produced 13.05 g of H₃PO₄, how many grams of hydrochloric acid would be need to start with?</em>
- Firstly, we should get the no. of moles (n) of 13.05 grams of H₃PO₄:
n = mass/molar mass = (13.05 g)/(97.994 g/mol) = 0.1332 mol.
<u><em>Using cross-multiplication:</em></u>
6.0 mol of HCl needed to produce → 2.0 mol of H₃PO₄, from stichiometry.
??? mol of HCl needed to produce → 0.1332 mol of H₃PO₄.
∴ The no. of moles of HCl needed = (6.0 mol)(0.1332 mol)/(2.0 mol) = 0.3995 mol.
∴ The mass of HCl needed = n*molar mass = (0.3995 mol)(36.46 g/mol) = 14.57 g.
<em>So, the grams of hydrochloric acid would be need to start with = 14.57 g.</em>
