Answer:
a)0.765 g
b)7.613 g
c)0.20 L
Explanation:
Consider the reaction involved;
Na(s) + HNO3(aq) ----> NaNO3(s) + H2(g)
Note that, if a hot, saturated aqueous solution of sodium nitrate was allowed to cool, solid sodium nitrate would crystallise out of the solution and this would also be classed as a precipitate. This is the case here.
Number of moles of sodium reacted= mass of sodium reacted/ molar mass of sodium
Number of moles of sodium= 7.85g/23gmol-1
Number of moles of sodium= 0.34 moles of sodium
Number of moles of acid reacted= concentration of acid × volume of acid
Number of moles of acid= 0.0450 × 200/1000
Number of moles of acid= 9×10^-3 moles
Therefore, HNO3 is the limiting reactant.
1 mole of HNO3 yield 1 mole of NaNO3
9×10^-3 moles of HNO3 yield 9×10^-3 moles of NaNO3
Hence mass of NaNO3= number of moles × molar mass
Mass of NaNO3= 9.0×10^-3 moles × 84.9947 g/mol
Mass of NaNO3= 0.765 g of NaNO3
b)
Since
1 mole of sodium metal reacts with 1 mole of HNO3
9×10^-3 moles of sodium reacts with 9×10^-3 moles of HNO3
Therefore amount of unreacted sodium metal = 0.34 moles - 9×10^-3 moles = 0.331 moles
Mass of unreacted sodium metal = 0.331 moles × 23 gmol-1= 7.613 g
c)
If 1 mole of HNO3 yields 1 mole of hydrogen gas
9×10^-3 moles of HNO3 yields 9×10^-3 moles of hydrogen gas.
1 mole of hydrogen gas occupies 22.4 L
9×10^-3 moles of hydrogen gas will occupy 9×10^-3 moles × 22.4/1 = 0.20 L
