The reaction of removing CO2
using LiOH is the following:
2 LiOH + CO2 -----> Li2CO3
+ H2O
By solving the amount of CO2
the LiOH can scrub:
(3.50 × 10^4 g LiOH) (1 mol LiOH/
24 g LiOH) ( 1 mol CO2 / 2 mol LiOH) ( 44 g CO2 /1 mol CO2) = 32, 083.33 g CO2
it can scrub
<span>Since number of astronaut = 32,
083.33 g / 9 (8.8 × 10^2) = 4 astronaut</span>
Here we have to choose the right option which tells the moles of CaCl₂ will react with 6.2 moles of AgNO₃ in the reaction
2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂
6.2 moles of silver nitrate (AgNO₃) will react with B. 3.1 moles of calcium chloride (CaCl₂).
From the reaction: 2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂
Thus 2 moles of AgNO₃ reacts with 1 mole of CaCl₂
Henceforth, 6.2 moles of AgNO₃ reacts with 
= 3.1 moles of CaCl₂.
1 mole of CaCl₂ reacts with 2 moles of AgNO₃. Thus-
A. 2.2 moles of CaCl₂ will react with 2.2×2 = 4.4 moles of AgNO₃.
C. 6.2 moles of CaCl₂ will reacts with 6.2×2 = 12.4 moles of AgNO₃.
D. 12.4 moles of CaCl₂ will reacts with 12.4 × 2 = 24.8 moles of AgNO₃
Thus the right answer is 6.2 moles of AgNO₃ will react with 3.1 moles of CaCl₂.
Answer: 66.2 g
Explanation:
1) The ratio of Al in the molecule is 1 mol to 1 mol .
2) The mass of 1 mol of molecules of Al (CH2H3O2)3 is the molar mass of the compound.
3) You calculate the molar mass of the compound using the atomic masses of each atom, in this way:
Al: 27 g/mol
C: 2 * 3 * 12 g/mol = 72 g/mol
H: 3 * 3 * 1 g/mol = 9 g/mol
O: 2 * 3 * 16 g/mol = 96 g/mol
Molar mass = 27 g/mol + 72 g/mol + 9 g/mol + 96 g/mol = 204 g/mol
4) Set a proportion:
27 g/mol x
-------------------- = ----------
204 g/mol 500 g
5) Solve for x:
x = 500 g * 27 g/mol / 204 g/mol = 66.2 g
An ester is the reaction product between A. alcohol and organic acid.
