Li2O (s) + H2O (g) →2LiOH (s) How many grams of Li2O must be carried on board to remove 70.0 kg of water?

What is the answer

Lithium oxide is used aboard the space shuttle to remove water from the atmosphere according to the equation

We need 116.1 kg of Li2O

Explanation:

Step 1: Data given

MAss of water = 70 kg = 70000 grams

Molar mass H2O = 18.02 g/mol

Molar mass of Li2O = 29.88 g/mol

Step 2: the balanced equation

Li2O (s) + H2O (g) →2LiOH (s)

Step 3: Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 70000 grams / 18.02 g/mol

Moles H2O = 3884.6 moles

Step 4: Calculate moles Li2O

For 1 mol Li2O we need 1 mol H2O to produce 2 moles LiOH

For 3884.6 moles H2O we need 3884.6 moles Li2O

Step 5: Calculate mass Li2O

MAss Li2O = moles * molar mass

MAss Li2O = 3884.6 moles * 29.88g/mol

Mass Li2O = 116071.8 grams = 116.1 kg

We need 116.1 kg of Li2O