Answer:
2NaCN + CaCO3 --> Na2CO3 + Ca(CN)2
Explanation:
Knowing the names gets us: NaCN + CaCO3 --> Na2CO3 + Ca(CN)2
Balance: there are two sodiums and cyanides on the product side so add a 2 to the reactant side.
Answer:
![[SO_3]=0.25M](https://tex.z-dn.net/?f=%5BSO_3%5D%3D0.25M)
Explanation:
Hello there!
In this case, since the integrated rate law for a second-order reaction is:
![[SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B%5BSO_3%5D_0%7D%7B1%2Bkt%5BSO_3%5D_0%7D)
Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:
![[SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B1.44M%7D%7B1%2B14.1M%5E%7B-1%7Ds%5E%7B-1%7D%2A0.240s%2A1.44M%7D%5C%5C%5C%5C)
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Balanced chemical equation:
2 C2H2 + 5 O2 = 4 CO2 + 2 H2O
2 moles C2H2 ---------------- 5 moles O2
moles C2H2 ------------------ 84 moles O2
moles C2H2 = 84 * 2 / 5
molesC2H2 = 168 / 5 => 33.6 moles of C2H2
