If you have a magnesium for every oxygen, then you have to start with two magnesiums. So the balanced equation is 2 Mg + O2 2 MgO.
And
CaCO3———→CaO + CO2
I hope it helped!
Answer:
8:10
Explanation:
The coefficients of a balanced chemical equation give us the mole ratios. The coefficient of carbon dioxide here is 8 and water is 10.
The Balanced Chemical Equation is as follow;
4 KO₂ + 2 CO₂ → 2 K₂CO₃ + 3 O₂
First find out the Limiting Reagent,
According to equation,
284 g (4 moles) KO₂ reacted with = 44.8 L (2 moles) of CO₂
So,
27.9 g of KO₂ will react with = X L of CO₂
Solving for X,
X = (44.8 L × 27.9 g) ÷ 284 g
X = 4.40 L of CO₂
Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,
According to eq.
284 g (4 moles) KO₂ formed = 138.2 g of K₂CO₃
So,
27.9 g of KO₂ will form = X g of K₂CO₃
Solving for X,
X = (138.2 g × 27.9 g) ÷ 284 g
X = 13.57 g of K₂CO₃
So, 13.57 g of K₂CO₃ formed is the theoretical yield.
%age Yield = 13.57 / 21.8 × 100
%age Yield = 62.24 %
<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399
<u>Explanation:</u>
Mass defect is defined as the difference in the mass of an isotope and its mass number.
The equation used to calculate mass defect follows:
![\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M](https://tex.z-dn.net/?f=%5CDelta%20m%3D%5B%28n_p%5Ctimes%20m_p%29%2B%28n_n%5Ctimes%20m_n%29%5D-M)
where,
= number of protons
= mass of one proton
= number of neutrons
= mass of one neutron
M = mass number of element
We are given:
An isotope of phosphorus which is 
Number of protons = atomic number = 15
Number of neutrons = Mass number - atomic number = 31 - 15 = 16
Mass of proton = 1.00728 amu
Mass of neutron = 1.00866 amu
Mass number of phosphorus = 30.973765 amu
Putting values in above equation, we get:
![\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399](https://tex.z-dn.net/?f=%5CDelta%20m%3D%5B%2815%5Ctimes%201.00728%29%2B%2816%5Ctimes%201.00866%29%5D-30.973765%5C%5C%5C%5C%5CDelta%20m%3D0.27399)
Hence, the mass defect for the formation of phosphorus-31 is 0.27399
Answer : The number of moles present in ammonia is, 70.459 moles.
Solution : Given,
Mass of ammonia = 
Molar mass of ammonia = 17.031 g/mole
Formula used :
Therefore, the number of moles present in ammonia is, 70.459 moles.
