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?\n0.375 m\n1.50 m\n3.00 m\n6.00 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?\n0.375 m\n1.50 m\n3.00 m\n6.00 m

Answer

Explanation:

Step1: Calculate moles of $Ca(OH)_2$

Use the formula $n = M\times V$, where $M$ is molarity and $V$ is volume. Given $M_{Ca(OH)2}=1.50\ M$ and $V{Ca(OH)2}=2.00\ L$. So $n{Ca(OH)_2}=1.50\ mol/L\times2.00\ L = 3.00\ mol$.

Step2: Determine moles of $HCl$ from mole - ratio

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

Step3: Calculate molarity of $HCl$

Use the formula $M=\frac{n}{V}$. Given $n_{HCl}=6.00\ mol$ and $V_{HCl}=1.00\ L$. So $M_{HCl}=\frac{6.00\ mol}{1.00\ L}=6.00\ M$.

Answer:

$6.00\ M$