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3 years ago

WHAT MASS OF WATER WILL BE PRODUCED FROM 2.70 MOLES OF CA(OH)2 REACTING WITH HCI

Chemistry

1 answer:

Keith_Richards [23]3 years ago

3 0

Answer: The mass of water produced in the reaction is 97.2 grams

Explanation:

We are given:

Moles of calcium hydroxide = 2.70 moles

The chemical equation for the reaction of calcium hydroxide and HCl follows:

$Ca(OH)_2+HCl\rightarrow CaCl_2+2H_2O$

By Stoichiometry of the reaction:

1 mole of calcium hydroxide produces 2 moles of water

So, 2.70 moles of calcium hydroxide will produce = $\frac{2}{1}\times 2.70=5.40mol$ of HCl

To calculate mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of water = 18 g/mol

Moles of water = 5.40 moles

Putting values in above equation, we get:

$5.40mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(5.40mol\times 18g/mol)=97.2g$

Hence, the mass of water produced in the reaction is 97.2 grams

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Answer:

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3 years ago

A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

$-w=\frac{dm}{dt}$

w is the mass flow

m is the mass of salt

$-v*C=\frac{dm}{dt}$

v is the volume flow

C is the concentration

$C=\frac{m}{V+(6-3)*L/min*t}$

$-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}$

$-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}$

$-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)$

$-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)$

$m=90kg*[2000L/(2000L+3*L/min*t)]$

a) Initially: t=0

$m=90kg*[2000L/(2000L+3*L/min*0)]=90kg$

b) t=210 min (3.5 hr)

$m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg$

c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

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3 years ago