The mass of oxygen gas is released in the reaction of decomposition of magnesium oxide 6.4 grams
calculation
mass of oxygen=mass of magnesium oxide -mass of magnesium produced
that is 16.12-9.72=6.4 grams
Answer:
34.81 ×10²³ atoms
Explanation:
Given data:
Mass of water = 52.0 g
Number of atoms of hydrogen = ?
Solution:
Number of moles of water:
Number of moles = mass/molar mass
Number of moles = 52.0 g/ 18 g/mol
Number of moles = 2.89 mol
1 mole of water contain 2 mole of hydrogen.
2.89 mole of water contain 2.89× 2= 5.78 moles of hydrogen.
Number of atoms of hydrogen:
1 mole = 6.022×10²³ atoms
5.78 mol × 6.022×10²³ atoms / 1mol
34.81 ×10²³ atoms
Answer:
Given info : 500 cc of 2N Na2CO3 are mixed with 400 cc of 3N H2SO4 and volume was diluted to one litre. To find : will the resulting solution is acidic , basic or neutral ? Calculate the molarity of the dilute solution. solution : no of moles of Na2CO3 = normality/n %3D - factor x volume 2/2 x 500/1000 = 0.5 mol %D no of moles of H2SO4 = 3/2 x 400/100O = 0.6 mol %3D We see, Na2CO3 + H2S04 => Na2S04 + CO2 + H2O Here one mol of Na2C03 reacts with one mole of H2SO4. So, 0.5 mol of Na2CO3 reacts with 0.5 mol of H2SO4. so, remaining 0.1 mol of H2SO4 makes solution acidic. Now molarity of solution = remaining no of moles of H2SO4/volume of solution= 0.1/1 = %3D 0.1M
Answer:
3.29 L
Explanation:
Use Charle's Law and rearrange formula.
- Hope that helped! Please let me know if you need further explanation.
Answer: 51.45 grams of excess reagent is left after the completion of reaction.
Explanation: For the calculation of moles, we use the formula:
....(1)
Given mass = 92 grams
Molar mass = 28g/mol
Putting values in equation 1, we get:
![Moles=\frac{92g}{28g/mol}=3.285moles](https://tex.z-dn.net/?f=Moles%3D%5Cfrac%7B92g%7D%7B28g%2Fmol%7D%3D3.285moles)
- For
![Cr_2O_3](https://tex.z-dn.net/?f=Cr_2O_3)
Given mass = 112 grams
Molar mass = 116g/mol
Putting values in equation 1, we get:
![Moles=\frac{112g}{116g/mol}=0.965moles](https://tex.z-dn.net/?f=Moles%3D%5Cfrac%7B112g%7D%7B116g%2Fmol%7D%3D0.965moles)
The reaction follows:
![2Cr_2O_3(s)+3Si(s)\rightarrow 4Cr(l)+3SiO_2(s)](https://tex.z-dn.net/?f=2Cr_2O_3%28s%29%2B3Si%28s%29%5Crightarrow%204Cr%28l%29%2B3SiO_2%28s%29)
By Stoichiometry,
2 moles of
reacts with 3 moles of silicon
So, 0.965 moles of
reacts with =
= 1.4475 moles of Silicon.
As, the moles of silicon is more than the required amount and is present in excess.
So, the excess reagent for the reaction is Silicon.
Moles of silicon remained after reaction = 3.285 - 1.4475 = 1.8375 moles
To calculate the amount of Silicon left in excess is calculated by using equation 1:
![1.8375=\frac{\text{Given mass}}{28}](https://tex.z-dn.net/?f=1.8375%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B28%7D)
Amount of Silicon in excess will be 51.45 grams.