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3 years ago

Find the number of moles of water that can be formed if you have 230 mol of hydrogen gas and 110 mol of oxygen gas

Chemistry

1 answer:

Verdich [7]3 years ago

3 0

Answer:

220mol.

Explanation:

Water is H2O. Hydrogen gas is H2. Oxygen gas is O2. You have 220mol of O and 460mol of H. O is the limiting reactant. The ratio O:H2O is 1:1. 220*1=220

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2 years ago

In the reaction 2 c o2 → 2 co, how many moles of carbon are needed to produce 66.0 g of carbon monoxide

weeeeeb [17]

The solution would be like this for this specific problem:

Given:

66.0 g of carbon monoxide

reaction 2 C + O2 → 2 CO

mol e= mass / molar mass 
mole of 2CO = 66.0g / (12.011 15.999)g / mol 
mole of 2CO = 2.36 (CO and C has a 1:1 mole ratio)

mole of 2CO = 2.36 -> mole of 1 CO = 2.36 / 2 = 1.18

We got 2 moles of C, thus 1.18 x 2 = 2.36

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3 years ago

Compared to the freezing point of 1.0 M KCl(aq) at standard pressure, the freezing point of 1.0 M CaCl2(aq) at standard pressure

Galina-37 [17]

Answer : The correct option is, (1) lower

Explanation :

Formula used for lowering in freezing point :

$\Delta T_f=i\times k_f\times m$

where,

$\DeltaT_b$ = change in freezing point

$k_b$ = freezing point constant

m = molality

i = Van't Hoff factor

According to the question, we conclude that the molality of the given solutions are the same. So, the freezing point depends only on the Van't Hoff factor.

Now we have to calculate the Van't Hoff factor for the given solutions.

(a) The dissociation of $KCl$ will be,

$KCl\rightarrow K^++Cl^-$

So, Van't Hoff factor = Number of solute particles = 1 + 1 = 2

(b) The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^-$

So, Van't Hoff factor = Number of solute particles = 1 + 2 = 3

The freezing point depends only on the Van't Hoff factor. That means higher the Van't Hoff factor, lower will be the freezing point and vice-versa.

Thus, compared to the freezing point of 1.0 M $KCl(aq)$ at standard pressure, the freezing point of 1.0 M $CaCl_2(aq)$ at standard pressure is lower.

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3 years ago