If 890 mL of 0.236 M aqueous NaI and 0.560 mol of liquid Br2 are reacted stoichiometrically according to the?
If 890 mL of 0.236 M aqueous NaI and 0.560 mol of liquid Br2 are reacted stoichiometrically according to the balanced equation, how many moles of liquid Br2 remain? Round your answer to 3 significant figures.
2NaI(aq) + Br2(l) ¡æ 2NaBr(aq) + I2(s) If 890 mL of 0.236 M aqueous NaI and 0.560 mol of liquid Br2 are reacted stoichiometrically according to the balanced equation, how many moles of liquid Br2 remain? Round your answer to 3 significant figures.
2NaI(aq) + Br2(l) ¡æ 2NaBr(aq) + I2(s)
Molar Mass (g/mol)
NaI 149.89
Br2 159.81
Density (g/mL)
Br2 3.12
Molar Volume (L)
22.4 at STP
Gas Constant
(L.atm.mol-1.K-1)
0.0821
I got
Br2 left = 0.56-(0.89x0.236)/2
=.45498
is this right??
Yes but the answer was to be rounded to 3 s.f. Round to 0.455 moles.
To determine the number of moles of liquid Br2 that remain after the reaction, we first need to calculate the moles of NaI used in the reaction.
Given:
Volume of NaI solution = 890 mL = 0.89 L
Molarity of NaI solution = 0.236 M
Using the equation Molarity = Moles/Volume, we can calculate the moles of NaI:
Moles of NaI = Molarity x Volume
Moles of NaI = 0.236 M x 0.89 L = 0.209 mol NaI
Now, according to the balanced equation, the stoichiometric ratio between NaI and Br2 is 2:1. This means that 2 moles of NaI react with 1 mole of Br2.
Since the reaction is stoichiometric, the moles of Br2 consumed will be equal to half the moles of NaI used.
Moles of Br2 consumed = 0.209 mol NaI / 2 = 0.105 mol Br2
However, we have 0.560 mol of Br2 initially. Therefore, the moles of Br2 remaining can be calculated by subtracting the moles of Br2 consumed from the initial moles of Br2.
Moles of Br2 remaining = 0.560 mol Br2 - 0.105 mol Br2 = 0.455 mol Br2
So, the correct answer is approximately 0.455 mol of liquid Br2 remaining after the reaction.
Your calculation is almost correct, but you made a small error in the calculation. Instead of subtracting (0.89 x 0.236) from 0.56, you should divide it by 2. Therefore, the correct calculation should be:
Br2 remaining = 0.56 - (0.89 x 0.236) / 2
Br2 remaining = 0.56 - (0.20984) / 2
Br2 remaining ≈ 0.56 - 0.10492
Br2 remaining ≈ 0.455
So, your answer of 0.455 mol for Br2 remaining is correct. Well done!