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katovenus [111]
3 years ago
10

Balancing equations

Chemistry
1 answer:
Evgesh-ka [11]3 years ago
5 0
No 11) one mole of Zn and two mole of HCL to get one mole of ZnCl and one mole of H2.
no 12) one mole of Mg and two mole of AgNO3 to get one mole of Mg(No3)2 and two mole of Ag.
no 13) one mole of CaCl2 and two mole AgNO3 to get two mole of AgCl and one mole of Ca(NO3)2.
no 14) two mole of Na3PO4 and three mole of BaCL2 to get one mole of Ba3(PO4)2 and six mole of NaCl.
no 15) two mole of NH4Cl and one mole of Ba(OH)2 to get one mole of BaCL2,two mole of NH3 and two mole of H2O.(am not much sure about this).
no 16) one mole of Na2CO3 and two mole of HCL to get two mole of NaCl,one mole of H2O and one mole of CO2.
no 17) one mole of K3PO4 and three mole of AgNO3 to get three mole of KNO3 and one mole of AgPO4.
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This is an example of Charles’ Law problems, the basic equation is: V1/T1 = V2/T2. One vital thing to recall for all gas law problems is that the temperature must be in Kelvin (not Celsius). 
So our given is 10.0 C = 283 K. So 
V1/T1 = V2/T2 
733/283 = 950/T2 

T2 = 367 K 
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If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc
hodyreva [135]

Answer: 69.72 kg of cryolite will be produced.

Explanation:

The balanced chemical equation is:

Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of Al_2O_3 = \frac{16.9\times 1000g}{102g/mol}=165.7moles

moles of NaOH = \frac{57.4\times 1000g}{40g/mol}=1435moles

moles of HF = \frac{57.4\times 1000g}{20g/mol}=2870moles

As 1 mole of Al_2O_3 reacts with 6 moles of NaOH

166 moles of  Al_2O_3 reacts with = \frac{6}{1}\times 166=996 moles of NaOH

As 1 mole of Al_2O_3 reacts with 12 moles of HF

166 moles of  Al_2O_3 reacts with = \frac{12}{1}\times 166=1992 moles of HF

Thus Al_2O_3 is the limiting reagent.

As 1 mole of Al_2O_3 produces = 2 moles of cryolite

166 moles of  Al_2O_3 reacts with = \frac{2}{1}\times 166=332 moles of cryolite

Mass of cryolite (Na_3AlF_6) = moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg

Thus 69.72 kg of cryolite will be produced.

8 0
3 years ago
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