if 19.1 mL of 1.26 M HBr is headed to 23.7g of CAC03 what volume of CO2 would be produced as 713.6 torr and 29.9°C

0.32 L

Explanation:

The equation of the reaction is;

2HBr(aq) + CaCO3(s) >CaBr2(aq) + H2O(l) + CO2(g)

The number of moles HBr reacted = 19.1/1000 L * 1.26 M = 0.024 moles

According to the reaction equation;

2 moles of HBr produces 1 mole of CO2

0.024 moles of HBr produces  0.024 * 1/2 = 0.012 moles of CO2

For CaCO3

Number of moles  = 23.7g/100 g/mol = 0.237 moles

Since the reaction is 1:1, 0.237 moles of CO2 is produced.

Hence HBr is the limiting reactant and 0.012 moles of CO2 is produced.

From PV =nRT

V = nRT/P

P= 713.6 torr

n= 0.012 moles

T = 29.9°C + 273 = 302.9 K

R =  62.36 L torr/mol K

V = 0.012 * 62.36 * 302.9/713.6

V = 0.32 L

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2