Answer:
Balancing chemical equation means making a number of atoms or molecules equal on both sides. In other words, this means that the number of atoms and molecules of each reacting element needs to be the same as the number of atoms and molecules of those elements in the product.
Our reaction is:
AlBr3 + K2SO4 -> KBr + Al2(SO4)3
and we need to balance it.
Since there are 3 molecules of SO4 in the product we need to put 3 before the reactant K2SO4. There are also 2 atoms of Al in the product, so we need to put 2 in front AlBr3. Now we have 6 atoms of K and Br on the left side, so we need to put 6 in front of KBr in the product.
So, our balanced equation will look like this:
2AlBr3 + 3K2SO4 -> 6KBr + Al2(SO4)3
4NH3 + 5O2 ==> 4NO + 6H2O Balanced equation
ALWAYS WORK IN MOLES, NOT IN GRAMS
moles of NO produced = 70.5 g NO x 1 mole/30 g = 2.35 moles NO
Since this represents only a 29.8% yield, find what 100% yield would be:
2.35 moles/0.298 = 7.89 moles of NO
From the balanced equation 4 moles NH3 produces 4 moles of NO. Calculate moles of NH3 needed:
7.89 moles NO x 4 moles NH3/4 moles NO = 7.89 moles NH3 needed
Find grams of NH3 needed:
7.89 moles NH3 x 17 g/mole = 134 g NH3 needed
Answer:
378mL
Explanation:
The following data were obtained from the question:
Pressure (P) = 99.19 kPa
Temperature (T) = 28°C
Number of mole (n) = 0.015 mole
Volume (V) =...?
Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:
For Pressure:
101.325 KPa = 1 atm
Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm
For Temperature:
T(K) = T(°C) + 273
T(°C) = 28°C
T(K) = 28°C + 273 = 301K.
Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:
PV = nRT
Pressure (P) = 0.98 atm
Temperature (T) = 301K
Number of mole (n) = 0.015 mole
Gas constant (R) = 0.0821atm.L/Kmol.
Volume (V) =...?
0.98 x V = 0.015 x 0.0821 x 301
Divide both side by 0.98
V = (0.015 x 0.0821 x 301) /0.98
V = 0.378 L
Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:
1L = 1000mL
Therefore, 0.378L = 0.378 x 1000 = 378mL
Therefore, the volume of N2 collected is 378mL
The standard temperature is 0c.
the standard pressure is 1atm.
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