Consider the reaction 2Al + 6HBr → 2AlBr3 + 3H2. If 8 moles of Al react with 8 moles of HBr, what is the limiting reactant?
1 answer:
Answer:- HBr is limiting reactant.
Solution:- The given balanced equation is:
From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.
The calculations could be shown as:
= 24 mol HBr
From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.
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Answer:- 0.273 kg
Solution:- A double replacement reaction takes place. The balanced equation is:
We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.
= 63.8 g aluminum nitrate
From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.
We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.
=
sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.
= 0.273 kg
So, 0.273 kg of 35% m/m sodium chlorate solution are required.