Question

What volume (in mL) of 0.250 M HNO3 reacts with 34.9 mL of 0.150 M Na2CO3 in the following reaction? 2 HNO3(aq) + Na2CO3(aq) --> 2 NaNO3(aq) + H2O(l) + CO2(g)

Answer 1

2 HNO3(aq) + Na2CO3(aq) --> 2 NaNO3(aq) + H2O(l) + CO2(g)

moles of Na2CO3 = 0.150 x 34.9 /1000 = 0.005235

2 x 0.005235 = 0.01047 moles HNO3 required

0.01047 = 0.250 x V

V = 0.04188 L

V = 41.88 mL

volume of HNO3 required = 41.88 mL