a 1.00 l volume of hcl reacted completely with 2.00 l of 1.50 m ca(oh)2 according to the balanced chemical…

a 1.00 l volume of hcl reacted completely with 2.00 l of 1.50 m ca(oh)2 according to the balanced chemical equation below.\n2hcl + ca(oh)2 → cacl2 + 2h2o\nwhat was the molarity of the hcl solution?\no 0.375 m\no 1.50 m\no 3.00 m\no 6.00 m

a 1.00 l volume of hcl reacted completely with 2.00 l of 1.50 m ca(oh)2 according to the balanced chemical equation below.\n2hcl + ca(oh)2 → cacl2 + 2h2o\nwhat was the molarity of the hcl solution?\no 0.375 m\no 1.50 m\no 3.00 m\no 6.00 m

Answer

Answer:

D. 6.00 M

Explanation:

Step1: Calculate moles of Ca(OH)₂

$n = M\times V$, where $M = 1.50\ M$ and $V=2.00\ L$. So $n_{Ca(OH)_2}=1.50\ mol/L\times2.00\ L = 3.00\ mol$.

Step2: Use mole - ratio from balanced equation

From $2HCl + Ca(OH)2\rightarrow CaCl_2 + 2H_2O$, the mole - ratio of $HCl$ to $Ca(OH)2$ is $2:1$. So $n{HCl}=2\times n{Ca(OH)_2}=2\times3.00\ mol = 6.00\ mol$.

Step3: Calculate molarity of HCl

$M=\frac{n}{V}$, with $n = 6.00\ mol$ and $V = 1.00\ L$. So $M_{HCl}=\frac{6.00\ mol}{1.00\ L}=6.00\ M$.