How many ml of 0.563 m hno3 are needed to dissolve 7.83 g of bacoz? 2hno3(aq) + baco3(s) — ba(no3)2(aq) + h2o(1) + co2(8) ml

Volume of HNO₃ required = 140 mL

Explanation:

Given data:

Molarity of HNO₃ = 0.563 M

mass of BaCO₃ = 7.83 g

Volume of HNO₃ = ?

Solution:

First of all we will write the balance chemical equation

2HNO₃ +  BaCO₃   →    Ba(NO₃)₂  + H₂O + CO₂

Number of moles of BaCO₃ = mass /  molar mass

Number of moles of BaCO₃ = 7.83 g / 197.34 g/mol

Number of moles of BaCO₃ = 0.04 mol

Now we compare the moles of  BaCO₃  and HNO₃ .

                   BaCO₃       :     HNO₃  

                      1              :       2

                     0.04       :       2×0.04 = 0.08 mol

Volume of HNO₃ required = number of moles / Molarity

Volume of HNO₃ required =  0.08 mol / 0.563 mol/L

Volume of HNO₃ required =  0.14 L

0.14 × 1000 = 140 mL

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2