A mole is a very important unit of measurement that chemists use. A mole of something means you have 602,214,076,000,000,000,000,000 of that thing, like how having a dozen eggs means you have twelve eggs.

$1 mol Cu + 2 mol AgNO_3$ → $2 mol Ag + 1 mol Cu(NO3)2$

63.55 g Cu —> 2 x 107.688 g Ag

63.55 g Cu gives 215.376 g of Ag

So, 49.1 g Cu —> $\frac{215.376 g X 49.1 g}{63.55 g}$

= 166.4 g Ag

Hence, 166.4 g Ag grams of silver can be produced from 49.1 g of copper.

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5 0

1 year ago

What Most Often Causes The Availability Of Water To Change?

Margaret [11]

Answer:the answer is C

Explanation:

8 0

2 years ago

One of the reactions in a blast furnace used to reduce iron is shown above. How many grams of Fe2O3 are required to produce 15.5

Salsk061 [2.6K]

Answer : The correct option is, (b) 22.1 g

Solution : Given,

Mass of iron = 15.5 g

Molar mass of iron = 56 g/mole

Molar mass of $Fe_2O_3$ = 160 g/mole

First we have to calculate the moles of iron.

$\text{Moles of Fe}=\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}}=\frac{15.5g}{56g/mole}=0.276moles$

Now we have to calculate the moles of $Fe_2O_3$ .

The balanced reaction is,

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

From the balanced reaction, we conclude that

As, 2 moles of iron obtained from 1 mole of $Fe_2O_3$

So, 0.276 moles of iron obtained from $\frac{0.276}{2}=0.138$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{Mass of }Fe_2O_3=\text{Moles of }Fe_2O_3\times \text{Molar mass of }Fe_2O_3$

$\text{Mass of }Fe_2O_3=(0.138mole)\times (160g/mole)=22.08g=22.1g$

Therefore, the amount of $Fe_2O_3$ required are, 22.1 grams.

6 0

3 years ago