Question

if you wanted to make 36.0 grams of h2o (water) using hydrogen (h2) and oxygen (o2) gas, how many grams of oxygen (o2) would you need? stoichiometric calculations equation: 2 h2(g) + o2(g) → 2 h2o(l) molecules: 2 molecules h2 + 1 molecule o2 → 2 molecules h2o mass (amu): 4.0 amu h2 + 32.0 amu o2 → 36.0 amu h2o amount (mol): 2 mol h2 + 1 mol o2 → 2 mol h2o mass (g): 4.0 g h2 + 32.0 g o2 → 36.0 g h2o the coefficients in the balanced equation give the ratio of moles of reactants and products 4.0 g o2 32.0 g o2 36.0 g o2

Explanation:

Step1: Analyze the stoichiometric table

From the given table, under the "Mass (g)" row, we see that to produce 36.0 g of \(H_2O\), the mass of \(O_2\) required is related to the reaction \(2H_2(g)+O_2(g)
ightarrow2H_2O(l)\). The table shows that 32.0 g of \(O_2\) reacts with 4.0 g of \(H_2\) to form 36.0 g of \(H_2O\).

Step2: Match with the problem

The problem asks for the mass of \(O_2\) needed to make 36.0 g of \(H_2O\). From the table, when the mass of \(H_2O\) is 36.0 g, the mass of \(O_2\) is 32.0 g.

Answer:

B. 32.0 g O2