The balanced chemical equation would be as follows:
<span>NaCl + AgNO3 -> NaNO3 + AgCl
We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:
0.0440 mol/L NaCl (.025 L) = 0.0011 mol NaCl -----> consumed completely and therefore the limiting reactant
0.320 mol/L AgNO3 (0.025 L) = 0.008 mol AgNO3
0.0011 mol NaCl ( 1 mol AgCl / 1 mol NaCl) = 0.0011 AgCl precipitate produced
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This answer is C hope this helped
Explanation:
Below is an attachment containing the solution
Answer:
227 mL KBr
Explanation:
To find the amount of milliliters KBr, you need to (1) convert grams to moles (via molar mass from values on periodic table), then (2) find the amount of liters KBr (via molarity equation using molarity and moles), and then (3) convert liters to milliliters. The final answer should have 3 sig figs to match the amount of sig figs in the given values.
<u>(Step 1)</u>
Molar Mass (KBr): 39.098 g/mol + 79.904 g/mol
Molar Mass (KBr): 119.002 g/mol
17.2 grams KBr 1 mole
----------------------- x ------------------ = 0.145 moles KBr
119.002 g
<u>(Step 2)</u>
Molarity (M) = moles / volume (L)
0.640 M = 0.145 moles / volume
(0.640 M) x (volume) = 0.145 moles
volume = (0.145 moles) / (0.640 M)
volume = 0.227 L
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<u>(Step 3)</u>
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0.227 L KBr 1,000 mL
------------------ x ----------------- = 227 mL KBr
1 L
