To determine the amount of HCl that reacts completely with 15 g of CaCO3, we can use the stoichiometry of the reaction:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
The first step is to calculate the molar mass of CaCO3 and HCl:
Next, find the moles of CaCO3 in 15 g:
Moles of CaCO3 = 15 g100 g/mol = 0.15 moles
According to the balanced equation, 1 mole of CaCO3 reacts with 2 moles of HCl. Thus, 0.15 moles of CaCO3 will require:
0.15 moles of CaCO3 × 2 moles of HCl/mole of CaCO3 = 0.30 moles of HCl
Now, convert moles of HCl to grams:
Mass of HCl = 0.30 moles × 36.5 g/mol = 10.95 g
Therefore, 15 g of CaCO3 completely reacts with 10.95 g of HCl.
The reaction between calcium carbonate (\( \text{CaCO}_3 \)) and hydrochloric acid (HCl) is given by the following equation: \[ \text{CaCO}_3 + 2 \, \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \]
From the balanced equation, it is clear that 1 mole of \( \text{CaCO}_3 \) reacts with 2 moles of HCl.
Now, let's calculate the moles of \( \text{CaCO}_3 \) in 15 g:
Molar mass of \( \text{CaCO}_3 \) = 40 + 12 + (3 \times 16) = 100 g/mol.
Moles of \( \text{CaCO}_3 \) = \( \frac{15 \, \text{g}}{100 \, \text{g/mol}} = 0.15 \, \text{mol} \).
From the equation, 1 mole of \( \text{CaCO}_3 \) reacts with 2 moles of HCl, so 0.15 moles of \( \text{CaCO}_3 \) will react with: \( 0.15 \, \text{mol} \times 2 = 0.30 \, \text{mol} \) of HCl. Now, calculate the mass of 0.30 moles of HCl:
Molar mass of HCl = 36.5 g/mol. Mass of HCl = \( 0.30 \, \text{mol} \times 36.5 \, \text{g/mol} = 10.95 \, \text{g} \).
Thus, the mass of HCl required is 10.95 g.
If 0.01 mol of $\mathrm{P_4O_{10}}$ is removed from 0.1 mol, then the remaining molecules of $\mathrm{P_4O_{10}}$ will be:
A solid cylinder of mass 2 kg and radius 0.2 m is rotating about its own axis without friction with angular velocity 5 rad/s. A particle of mass 1 kg moving with a velocity of 5 m/s strikes the cylinder and sticks to it as shown in figure.
The angular velocity of the system after the particle sticks to it will be: