Answer:
0.26g of NaCl is the maximum mass that could be produced
Explanation:
Based on the reaction:
HCl + NaOH → NaCl + H₂O
<em>Where 1 mol of HCl reacts per mol of NaOH to produce 1 mol of NaCl</em>
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To solve this question we need to find <em>limiting reactant. </em>The moles of limiting reactant = Moles of NaCl produced:
<em>Moles HCl -Molar mass: 36.46g/mol-:</em>
0.365g HCl * (1mol / 36.46g) = 0.010 moles HCl
<em>Moles NaOH -Molar mass: 40g/mol-:</em>
0.18g NaOH * (1mol / 40g) = 0.0045 moles NaOH
As the reaction is 1:1 and moles NaOH < moles HCl, limiting reactant is NaOH and maximum moles produced of NaCl are 0.0045 moles.
The mass of NaCl is:
<em>Mass NaCl -Molar mass: 58.44g/mol-:</em>
0.0045 moles * (58.44g/mol) =
<h3>0.26g of NaCl is the maximum mass that could be produced</h3>
Given: C3H8(g) + O2(g) ----> CO2 (g) + H2O (g)
Step : Put a 3 in front of CO2 (g) to balance C
=> C3H8(g) + O2(g) ----> 3CO2 + H2O to balance H
Step 2: Put a 4 in front of H2O
=> C3H8 (g) + O2(g) -----> 3CO2 (g) + 4H2O (g)
Step 3: Given that there are 3*2 + 4 = 10 O to the right side, put a 5 in front of O2 to balance O:
=> C3H8(g) + 5O2(g) -----> 3CO2(g) + 4H2O(g)
You can verify that the equation is balanced.
So, the answer is that the coefficient in front of O2 is 5.
