1. Which feature would most LIKELY be forming at converging continental tectonic plate

Use the following equation to answer the questions and please show all work.

DIA [1.3K]

Answer:

a.36 g of water is produced.

b.64 g of $O_{2}$ is consumed.

Explanation:

The reaction is $2H_{2} + O_{2}$ ⇒ $2H_{2}O$

a.

Given,

Weight of $H_{2}$ reacted = 4g

Weight of 1 mole of $H_{2}$ = 2 $\times$ 1 = 2g

Therefore no. of moles of $H_{2}$ reacted = $\frac{4}{2}$ = 2 moles;

Also given,

Weight of $O_{2}$ reacted = 32 g

Weight of 1 mole of $O_{2}$ = 2 $\times$ 16 = 32 g

Therefore no. of moles of $O_{2}$ reacted = $\frac{32}{32}$ = 1

We know that 2 moles of Hydrogen reacts with 1 mole of Oxygen to give 2 moles of water,

As we took 2 moles of Hydrogen and 1 mole of Oxygen,

Directly,from the equation we can tell 2moles of water will be produced.

Therefore no. of moles of $H_{2} O$ produced = 2

Weight of 1 mole of water = $2\times 1+16$ = 18

Therefore weight of $H_{2}O$ produced = $2\times 18$ = 36gm

b.

Given ,

72 g of $H_{2}O$ is produced.

So,

no. of moles of $H_{2}O$ produced = $\frac{72}{18}$ = 4 moles

From equation For every 2 moles of water formed , 1 mole of oxygen must be required.

So for producing 4 moles of water,

No. of moles of Oxygen required = 2 moles.

Therefore weight of $O_{2}$ reacted = $2\times32$ = 64 g

Method 2:

Given,

8 g of $H_{2}$ has reacted.

So,

no. of moles of $H_{2}$ reacted = $\frac{8}{2}$ = 4 moles.

From equation , we know that For every 2 moles of $H_{2}$ reacted,1 mole of $O_{2}$ will react.

Therefore,

No. of moles of $O_{2}$ that reacts with 4 moles of $H_{2}$ = $2\times1$ = 2 moles

Therefore the weight of $O_{2}$ reacted = $2\times 32$ = 64 g

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Hello!

BeH₂ is a linear molecule, while CaH₂ is an angular molecule.

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Have a nice day!

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