Question

1. 7.85 g of sodium metal is added to 200 mL of 0.0450 M HNO3

a) Predict the mass of the precipitate
b) Predict the amount of excess reagent left over
c) Volume of gaseous product at STP conditions

Answer 1

a)0.765 g

b)7.613 g

c)0.20 L

Step-by-step 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