Ca + 2HCl = CaCl₂ + H₂

c=4.50 mol/l
v=2.20 l

n(HCl)=cv

m(Ca)/M(Ca)=n(HCl)/2

m(Ca)=M(Ca)cv/2

m(Ca)=40g/mol·4.50mol/l·2.20l/2=198 g

198 grams of Ca are needed

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3 years ago

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How much heat is required to convert 20.0 g of ice at 50.0⁰C to liquid water at 0.0⁰C? The specific heat of ice is 2.06 J/(g∙⁰C)

Alex777 [14]

Answer:

8740 joules are required to convert 20 grams of ice to liquid water.

Explanation:

The amount of heat required ( $Q$ ), measured in joules, to convert ice at -50.0 ºC to liquid water at 0.0 ºC is the sum of sensible heat associated with ice and latent heat of fussion. That is:

$Q = m\cdot [c\cdot (T_{f}-T_{o})+L_{f}]$ (1)

Where:

$m$ - Mass, measured in grams.

$c$ - Specific heat of ice, measured in joules per gram-degree Celsius.

$T_{o}$ , $T_{f}$ - Temperature, measured in degrees Celsius.

$L_{f}$ - Latent heat of fussion, measured in joules per gram.

If we know that $m = 20\,g$ , $c = 2.06\,\frac{J}{g\cdot ^{\circ}C}$ , $T_{f} = 0\,^{\circ}C$ , $T_{o} = -50\,^{\circ}C$ and $L_{f} = 334\,\frac{J}{g }$ , then the amount of heat is: