Asked by AJ
What volume of 0.050 M AgNO3, silver nitrate solution is required to reat completely with 60.0 mL of 0.025 M K2S, potassium sulfide?
Answers
Answered by
Jai
First, we write the balanced reaction:
2 AgNO3 + K2S -> 2 KNO3 + Ag2S
Recall that molarity is moles per liter solution, or
M = n/V
Then we get the number of moles of K2S by simply multiplying the volume (should be in L units) by the concentration (Molarity has units mol/L):
n = V * M
n = 0.060 L * 0.025 M = 0.0015 mol K2S
From the balanced reaction, we get the ratio of AgNO3 and K2S. For every 1 mole of K2S, 2 moles of AgNO3 should be reacted. Therefore, to cancel units,
0.0015 mol K2S * (2 mol AgNO3 / 1 mol K2S) = 0.003 mol AgNO3
Now that we know the moles of AgNO3, to get its volume, we divide the moles by the concentration:
V = n / M
V = 0.003 / 0.05
V = 0.06 L
V = 60 mL AgNO3
Hope this helps~ `u`
2 AgNO3 + K2S -> 2 KNO3 + Ag2S
Recall that molarity is moles per liter solution, or
M = n/V
Then we get the number of moles of K2S by simply multiplying the volume (should be in L units) by the concentration (Molarity has units mol/L):
n = V * M
n = 0.060 L * 0.025 M = 0.0015 mol K2S
From the balanced reaction, we get the ratio of AgNO3 and K2S. For every 1 mole of K2S, 2 moles of AgNO3 should be reacted. Therefore, to cancel units,
0.0015 mol K2S * (2 mol AgNO3 / 1 mol K2S) = 0.003 mol AgNO3
Now that we know the moles of AgNO3, to get its volume, we divide the moles by the concentration:
V = n / M
V = 0.003 / 0.05
V = 0.06 L
V = 60 mL AgNO3
Hope this helps~ `u`
Answered by
AJ
Thank a bunch
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