Question

question 1
1 pts
consider the combustion of hydrogen for the next two questions.
$2h_2(g) + o_2(g) \longrightarrow +2h_2o(g)$
reaction is at standard temperature and pressure and all reagents are in the gas phase.
if we react 10 moles of hydrogen with 6 moles of oxygen. what is the limiting reagent and how much water is produced?

• oxygen and 6 moles water produced.
• oxygen and 5 moles water produced.
• hydrogen and 5 moles water produced.
• hydrogen and 10 moles water produced.

question 2
1 pts
how many grams of water are produced if we react 3 moles of hydrogen with 3 moles of oxygen?
$2h_2(g) + o_2(g) \longrightarrow +2h_2o(g)$

• about 40 grams
• about 80 grams
• about 20 grams
• about 60 grams

Explanation:

Question 1

Step1: Determine mole ratio from reaction

The reaction is \(2H_2(g)+O_2(g)
ightarrow2H_2O(g)\). The mole ratio of \(H_2:O_2:H_2O\) is \(2:1:2\).

Step2: Calculate moles of \(O_2\) needed for \(10\) moles \(H_2\)

From the ratio, moles of \(O_2\) needed for \(10\) moles \(H_2\) is \(\frac{1}{2}\times10 = 5\) moles. We have \(6\) moles of \(O_2\), so \(H_2\) will be used up first (limiting reagent).

Step3: Calculate moles of \(H_2O\) produced from \(H_2\)

Moles of \(H_2O\) produced is equal to moles of \(H_2\) (since ratio \(H_2:H_2O = 2:2 = 1:1\))? Wait, no, ratio \(H_2:H_2O\) is \(2:2\), so moles of \(H_2O=\frac{2}{2}\times10 = 10\)? Wait, no, wait: Wait, the ratio of \(H_2\) to \(H_2O\) is \(2:2\) (or \(1:1\)), so if \(10\) moles \(H_2\) react, moles of \(H_2O\) is \(10\)? Wait, no, let's re - check. The reaction is \(2H_2 + O_2
ightarrow2H_2O\). So for \(2\) moles \(H_2\), we get \(2\) moles \(H_2O\). So for \(10\) moles \(H_2\), moles of \(H_2O=\frac{2}{2}\times10 = 10\) moles. And the moles of \(O_2\) required for \(10\) moles \(H_2\) is \(\frac{1}{2}\times10 = 5\) moles. We have \(6\) moles \(O_2\), so \(H_2\) is limiting. So the correct option is "Hydrogen and 10 moles water produced."

Step1: Determine limiting reagent

Reaction: \(2H_2(g)+O_2(g)
ightarrow2H_2O(g)\). Mole ratio \(H_2:O_2 = 2:1\). For \(3\) moles \(H_2\), moles of \(O_2\) needed is \(\frac{1}{2}\times3 = 1.5\) moles. We have \(3\) moles \(O_2\), so \(H_2\) is limiting.

Step2: Calculate moles of \(H_2O\) produced from \(H_2\)

Mole ratio \(H_2:H_2O = 2:2 = 1:1\), so moles of \(H_2O\) produced is \(3\) moles (since \(3\) moles \(H_2\) react).

Step3: Calculate mass of \(H_2O\)

Molar mass of \(H_2O\) is \(18\) g/mol. Mass = moles \(\times\) molar mass = \(3\times18 = 54\)? Wait, no, wait: Wait, the ratio of \(H_2\) to \(H_2O\) is \(2:2\), so for \(3\) moles \(H_2\), moles of \(H_2O=\frac{2}{2}\times3 = 3\) moles. Molar mass of \(H_2O\) is \(18\) g/mol. So mass is \(3\times18 = 54\), but the options are about 40, 80, 20, 60. Wait, maybe I made a mistake. Wait, let's re - calculate. Wait, the reaction is \(2H_2+O_2
ightarrow2H_2O\). So for \(3\) moles \(H_2\), moles of \(H_2O\) is \(3\) moles (since \(2\) moles \(H_2\) give \(2\) moles \(H_2O\)). Molar mass of \(H_2O\) is \(18\) g/mol. So \(3\times18 = 54\), which is about 60? Wait, no, maybe the question has a typo or my calculation is wrong. Wait, wait, maybe I messed up the limiting reagent. Wait, for \(3\) moles \(H_2\) and \(3\) moles \(O_2\). The moles of \(O_2\) needed for \(3\) moles \(H_2\) is \(1.5\) moles. So \(H_2\) is limiting. Moles of \(H_2O\) is \(3\) moles. Mass is \(3\times18 = 54\), which is closest to 60? Wait, the options are "About 40 grams", "About 80 grams", "About 20 grams", "About 60 grams". So the answer is about 60 grams.

Answer:

Hydrogen and 10 moles water produced.

Question 2