Answer:
53.6 grams of silver chloride was produced.
Explanation:
![AgNO_3+HCl+\rightarrow AgCl+HNO_3](https://tex.z-dn.net/?f=AgNO_3%2BHCl%2B%5Crightarrow%20AgCl%2BHNO_3)
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
Mass of silver nitrate = 50.0 g
Mass of hydrogen chloride = 50.0 g
Mass of silver chloride = x
Mass of nitric acid = 46.4 g
Mass of silver nitrate + Mass of hydrogen chloride =
Mass of silver chloride + Mass of nitric acid
[te]50.0 g+50.0 g=x+46.4 g[/tex]
![x=50.0 g+50.0 g - 46.4 g = 53.6 g](https://tex.z-dn.net/?f=x%3D50.0%20g%2B50.0%20g%20-%2046.4%20g%20%3D%2053.6%20g)
53.6 grams of silver chloride was produced.
Data:
Molar Mass of NaOH = 40 g/mol
Solving: <span>According to the Law Avogradro, we have in 1 mole of a substance, 6.02x10²³ atoms/mol or molecules
</span>
1 mol -------------------- 6.02*10²³ molecules
y mol -------------------- 2.70*10²² molecules
6.02*10²³y = 0.270*10²³
![y = \frac{0.270*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}{6.02*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}](https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B0.270%2A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21%5C%2110%5E%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21%5C%2123%7D%7D%7B6.02%2A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21%5C%2110%5E%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21%5C%2123%7D%7D%20%20)
![\boxed{y \approx 0.04\:mol}](https://tex.z-dn.net/?f=%5Cboxed%7By%20%5Capprox%200.04%5C%3Amol%7D)
Solving: <span>Find the mass value now
</span>
40 g ----------------- 1 mol of NaOH
x g ------------- 0.04 mol of NaOH
![x = 40*0.04](https://tex.z-dn.net/?f=x%20%3D%2040%2A0.04)
![\boxed{\boxed{x = 1.6\:g}}\end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7Bx%20%3D%201.6%5C%3Ag%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
Answer:
The mass is 1.6 grams
1.1 Moles / 0.5 Liters = 0.22 Molarity
The reaction is
FeO + Fe3O4 + 1/2 O2---> 2Fe2O3
Thus as shown in the balanced equation two moles of Fe2O3 are formed when 0.5 moles of O2 reacted with mixture of FeO and Fe3O4
moles of Fe2O3 = MAss / Molar mass = 4.141 / 159.69 = 0.0259 moles
So moles of O2 needed = 0.5 X 0.0259 = 0.01295
Mass of O2 = moles X molar mass = 0.01295 X 32 = 0.4144 grams