Compound is X3Y2.
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Write a balance equation for the reaction between the analyte and the titrant.
Calculate the # of moles of titrant using the volume of titrant required and the concentration of titrant.
Calculate the # of moles of analyte using the stoichiometric coefficients of the equation.
Calculate the concentration of the analyte using the number or moles of analyte and the volume of analyte titrated.
Answer:
0.0583g
Explanation:
The equation of the reaction is;
2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)
From the question, number of moles of HNO3 reacted= concentration × volume
Concentration of HNO3= 0.100 M
Volume of HNO3 = 20.00mL
Number of moles of HNO3= 0.100 × 20/1000
Number of moles of HNO3 = 2×10^-3 moles
From the reaction equation;
2 moles of HNO3 reacts with 1 mole of Mg(OH)2
2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2
But
n= m/M
Where;
n= number of moles of Mg(OH)2
m= mass of Mg(OH)2
M= molar mass of Mg(OH)2
m= n×M
m= 1×10^-3 moles × 58.3 gmol-1
m = 0.0583g
Answer: 8moles
Explanation:
The reaction below shows the formation of 2 moles of water from 2 moles of hydrogen and 1 mole of oxygen respectively.
2H2(g) + O2 (g) --> 2H2O(l)
So, if 1 mole of O2 produce 2 mole of H2O
4 moles of O2 will produce Z mole of H2O
To get the value of Z, cross multiply
1 x Z = 4 x 2
Z = 8
So, the equation will be 8H2(g) + 4O2 (g) --> 8H2O(l)
Thus, 4 moles of O2 will produce 8moles of H2O .
