Question

How many mL of 0.577 M HNO3 are needed to dissolve 6.85 g of Baco,? 2HNO3(aq) + BaCO3(s) — Ba(NO3)2(aq) + H2O(l) + CO2(8) ml

Answer 1

Number of moles of BaCO3 = 6.85 g / 197.34 g/mol = 0.0347 mole

From the balanced equation we can say that

1 mole of BaCO3 requires 2 mole of HNO3 so

0.0347 mole of BaCO3 will require

= 0.0347 mole of BaCO3 *(2 mole of HNO3 / 1 mole of BaCO3)

= 0.0694 mole of HNO3

molarity of HNO3 = number of moles of HNO3 / volume of solution in L

0.577 = 0.0694 / volume of solution in L

volume of solution in L = 0.0694 / 0.577 = 0.120 L

1 L = 1000 mL

0.120 L = 120 mL

Therefore, the volume of HNO3 required would be 120 mL