Answer:
11.3 g.
Explanation:
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In this case, since the combustion of butane is:
Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:
Therefore, the resulting mass of water is:
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Answer: B. 4 moles Fe and 3 moles CO2
Explanation:
write a balanced chemical reaction
that is FeO3 + 3 CO → 2 Fe + 3CO2
2 moles of Fe2O3 reacted with 3 x2=6 moles of Co to form Fe and CO2 therefore CO was in excess and Fe2O3 was limiting reagent.
use the mole ratio to determine the moles of each product.
that is the mole ratio 0f Fe2O3 : Fe is 1:2 therefore the moles of Fe = 2x2=4 moles
the mole ratio of Fe2CO3 : CO2 is 1: 3 therefore the moles of Co2 = 2 x3 = 6 moles
Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
