The balanced chemical reaction is written as:
<span>Zn(s) + 2HCl(aq) = ZnCl2(aq) + H2(g)
</span>
We are given the amount of the reactant zinc to be used in the reaction This amount would be the starting point for the calculation. It is as follows:
5.00 mol Zn ( 1 mol H2 / 1 mol Zn ) (2.02 g H2 / 1 mol H2 ) = 10.1 g H2
Therefore, the correct answer is the second option.
The first one is an alkane, the second is an alkene, and the third is an alkyne. The suffix of the compound tells you what kind of hydrocarbon it is.
<span>analogous and <span>homologous</span></span>
Answer:
22.77 g.
he limiting reactant is O₂, and the excess reactant is Mg.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (16.3 g) of Mg and (4.52 g) of oxygen:
no. of moles of Mg = mass/molar mass = (16.3 g)/(24.3 g/mol) = 0.6708 mol.
no. of moles of O₂ = mass/molar mass = (4.52 g)/(16.0 g/mol) = 0.2825 mol.
So. 0.565 mol of Mg reacts completely with (0.2825 mol) of O₂.
<em>∴ The limiting reactant is O₂, and the excess reactant is Mg (0.6708 - 0.565 = 0.1058 mol).</em>
<u><em>Using cross multiplication:</em></u>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.565 mol of Mg produce → <em>0.565 mol of MgO.</em>
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.565 mol)(40.3 g/mol) = <em>22.77 g.</em>
