6.0m(mol/kg) of HCl
125mL H2O = 0.125kg
6mol/kg = n mol/0.125kg, n = 0.75mol
When 0.75mol of HCl reacts, 0.75/2=0.375mol of H2 is produced. H2 = 2g/mol
So, 0.375mol H2 = 0.75g
The theoretical proportion is given by the balanced chemical equation:
2 mol NBr / 3 mol Na OH
Then x mol NaOH / 40 mol NBr3 = 3mol NaOH/2 mol NBr3
Solve for x, x = 40 * 3/2 = 60 mol NaOH.
Given that there are 48 mol NaOH (less than 60) this is the limitant reactant and the other is the excess reactant.
Answer: NBr3..
Magnesium(?)
<span>2 HCl + Mg ? MgCl2 + H2</span>
Answer: Hope this helps
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Answer:
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