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# An average person on board the International Space Station (ISS) requires 840 grams of oxygen per da

NY Regents Chemistry Exam June 2022

An average person on board the International Space Station (ISS) requires 840 grams of oxygen per day. To produce the oxygen needed on the ISS, water undergoes an electrolysis reaction. The oxygen produced is vented into the ISS cabin, and the hydrogen is vented into outer space. The reaction is represented by the balanced equation below.

2H2O(l) + energy → 2H2(g) + O2(g)

Some gases in the ISS must be removed from the air the astronauts breathe. Carbon dioxide can be removed using solid lithium hydroxide.

66 Show a numerical setup for calculating the number of moles of oxygen gas required for the average person per day. The gram-formula mass of O2(g) is 32 g/mol. 

Explanation: According to Table R on the Reference Table, number of moles = given mass/ gram formula mass. We are given 840 grams of oxygen. The gram-formula mass is given in the question.

Answer: number of moles = 840g/332g/mol.

67 State the change in oxidation number for oxygen during the electrolysis reaction represented by the equation. 

Explanation: Usually oxidation number of oxygen in a compound is -2, which can be seen on the Periodic Table (up right corner of the element box). When an element is on its own such as Fe, H2, O2...., the oxidation number is equal to zero.

68 Determine the number of moles of oxygen vented into the cabin when 120 moles of water undergoes electrolysis. 

Explanation: The coefficients in the equal can be used for stoichiometry calculation. For every 2 moles of H2O, we have 1 mole of O2 according to the equation. Therefore, for 120 mole of water, we would have half the amount of moles of oxygen.

69 Determine the percent composition by mass of hydrogen in water. 

Explanation: According to Table T on the Periodic Table, percent by mass = (mass of part/mass of whole) *100.

H2O consists of H (two atoms) and O (one atom). Masses of atoms of elements can be found in the Periodic Table

H = 2*1.00794 = 2.01588

O = 1* 15.9994 = 15.9994

Total mass of water (gram formula mass) = 2.01588+15.9994 = 18.01528 g/mol

percent by mass hydrogen = (2.01588/18.01528 )*100 = 11%

70 Balance the equation for the reaction between LiOH and CO2 in your answer booklet, using the smallest whole-number coefficients. 

Explanation: ___ LiOH + ___CO2 → ___Li2CO3 + ___ H2O We balance the equation by using coefficients only. Let's go element by element, doing oxygen last since we see oxygen in multiple place on the left and right sides. Starting with lithium, there is 1 lithium on the left and 2 lithiums on the right. We need to add 2 in front of LiOH on the left. We get:

2 LiOH + CO2 → Li2CO3 + H2O

Let's not look at hydrogen: 2 on the left (coefficient gets distributed) and 2 on the right = balanced

Carbon is 1 on the left and 1 on the right = balanced

Oxygen = 2+2 on the left and 3+1 on the right. 4=4 balanced!

The equation is balanced.

Answer: 2 LiOH + CO2 → Li2CO3 + H2O