To react with a $5.00 \cdot g$ $5.00g$ mass of calcium hydroxide, what volume of $0.100 \cdot m o l \cdot L^{- 1}$ $0.100mol*L^-1$ $H C l$ $HCl$ is required?

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To react with a $5.00 \cdot g$ $5.00g$ mass of calcium hydroxide, what volume of $0.100 \cdot m o l \cdot L^{- 1}$ $0.100mol*L^-1$ $H C l$ $HCl$ is required?

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Answer 1 · 2016-10-03T09:28:02

Approx. $1.4 \cdot L$ $1.4*L$ hydrochloric acid are required.

Explanation:

We need (i) a chemical reaction:

$C a {(O H)}_{2} (s) + 2 H C l (a q) \to C a C l_{2} (a q) + 2 H_{2} O (l)$ $Ca(OH)_2(s) + 2HCl(aq)rarr CaCl_2(aq) + 2H_2O(l)$ *

And (ii), the molar quantity of calcium hydroxide:

$=$ $=$ $\frac{5.00 \cdot g}{74.09 \cdot g \cdot m o l^{- 1}}$ $(5.00*g)/(74.09*g*mol^-1)$ $=$ $=$ $0.0675 \cdot m o l$ $0.0675*mol$ .

Given the rxn stoichiometry, clearly, we need $2 \times 0.0675 \cdot m o l = 0.135 \cdot m o l$ $2xx0.0675*mol=0.135*mol$ $H C l (a q)$ $HCl(aq)$ .

And given our concentration of $H C l$ $HCl$ , $\frac{0.135 \cdot m o l}{0.100 \cdot m o l \cdot L^{- 1}} \times 10^{3} \cdot m L \cdot L^{- 1}$ $(0.135*mol)/(0.100*mol*L^-1)xx10^3*mL*L^-1$ $=$ $=$ $? ? m L$ $??mL$

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To react with a $5.00 \cdot g$ $5.00g$ mass of calcium hydroxide, what volume of $0.100 \cdot m o l \cdot L^{- 1}$ $0.100mol*L^-1$ $H C l$ $HCl$ is required?

1 Answer

Explanation:

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To react with a 5.00⋅g5.00*g mass of calcium hydroxide, what volume of 0.100⋅mol⋅L−10.100*mol*L^-1 HClHCl is required?

1 Answer

Explanation:

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To react with a $5.00 \cdot g$ $5.00g$ mass of calcium hydroxide, what volume of $0.100 \cdot m o l \cdot L^{- 1}$ $0.100mol*L^-1$ $H C l$ $HCl$ is required?