Question

question 29 (1 point)
when 4.00 mol h₂ is mixed with 2.00 mol cl₂ how many moles of hcl can form?
h₂(g) + cl₂(g) → 2hcl
a) 2.00 mol hcl
b) 1.00 mol hcl
c) 4.00 mol hcl
d) 6.00 mol hcl

Explanation:

Step1: Determine the limiting reactant

The balanced chemical equation is \(H_{2}(g)+Cl_{2}(g)\to 2HCl\). The mole ratio of \(H_{2}\) to \(Cl_{2}\) is \(1:1\). We have \(n(H_{2}) = 4.00\ mol\) and \(n(Cl_{2})=2.00\ mol\). Since \(Cl_{2}\) has a smaller amount, \(Cl_{2}\) is the limiting reactant.

Step2: Calculate the moles of \(HCl\)

From the balanced equation, the mole ratio of \(Cl_{2}\) to \(HCl\) is \(1:2\). Using the formula \(n(HCl)=2\times n(Cl_{2})\) (based on stoichiometry). Substitute \(n(Cl_{2}) = 2.00\ mol\) into the formula: \(n(HCl)=2\times2.00\ mol\)

Answer:

a) \(2.00\ mol\ HCl\)